Q1. Find the derivative of the following functions from first principle:

1. \(-x\)
2. \((-x)^{-1}\)
3. \(\sin (x+1)\)
4. \(\cos (x-\frac{\pi}{8})\)

Solution

(i) Let the function be \( f(x) = -x \). Using the first principle of differentiation, the derivative is defined as

\[ \begin{aligned} f'(x) &= \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\ &= \lim_{h \to 0} \frac{-(x+h) - (-x)}{h} \\ &= \lim_{h \to 0} \frac{-h}{h} \\ &= -1 \end{aligned} \]

Hence, the derivative of \(-x\) is \(-1\).

(ii) Let the function be \( f(x) = (-x)^{-1} = -\dfrac{1}{x} \). Applying the first principle,

\[ \begin{aligned} f'(x) &= \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\ &= \lim_{h \to 0} \frac{-\dfrac{1}{x+h} + \dfrac{1}{x}}{h} \\ &= \lim_{h \to 0} \frac{-x + (x+h)}{h x (x+h)} \\ &= \lim_{h \to 0} \frac{h}{h x (x+h)} \\ &= \frac{1}{x^2} \end{aligned} \]

Therefore, the derivative of \((-x)^{-1}\) is \(\dfrac{1}{x^2}\).

(iii) Let the function be \( f(x) = \sin(x+1) \). Using the first principle,

\[ \begin{aligned} f'(x) &= \lim_{h \to 0} \frac{\sin(x+1+h) - \sin(x+1)}{h} \\ &= \lim_{h \to 0} \frac{2 \cos\left(\dfrac{2x+2+h}{2}\right) \sin\left(\dfrac{h}{2}\right)}{h} \\ &= \lim_{h \to 0} \cos\left(x+1+\dfrac{h}{2}\right) \cdot \frac{\sin\left(\dfrac{h}{2}\right)}{\dfrac{h}{2}} \\ &= \cos(x+1) \end{aligned} \]

This follows from the standard limit \(\lim_{u \to 0} \dfrac{\sin u}{u} = 1\).

(iv) Let the function be \( f(x) = \cos\left(x - \dfrac{\pi}{8}\right) \). Using the first principle,

\[ \begin{aligned} f'(x) &= \lim_{h \to 0} \frac{\cos\left(x - \dfrac{\pi}{8} + h\right) - \cos\left(x - \dfrac{\pi}{8}\right)}{h} \\ &= \lim_{h \to 0} \frac{-2 \sin\left(x - \dfrac{\pi}{8} + \dfrac{h}{2}\right) \sin\left(\dfrac{h}{2}\right)}{h} \\ &= \lim_{h \to 0} -\sin\left(x - \dfrac{\pi}{8} + \dfrac{h}{2}\right) \cdot \frac{\sin\left(\dfrac{h}{2}\right)}{\dfrac{h}{2}} \\ &= -\sin\left(x - \dfrac{\pi}{8}\right) \end{aligned} \]

Thus, the derivative of \(\cos\left(x - \dfrac{\pi}{8}\right)\) is \(-\sin\left(x - \dfrac{\pi}{8}\right)\).

Q2. \(x+a\)

Solution

Let the given function be \( f(x) = x + a \), where \(a\) is a constant. To find the derivative, we differentiate each term of the function with respect to \(x\).

The derivative of \(x\) with respect to \(x\) is \(1\), and since \(a\) is a constant, its derivative is \(0\).

\[ \begin{aligned} f'(x) &= \dfrac{d}{dx}(x) + \dfrac{d}{dx}(a) \\ &= 1 + 0 \\ &= 1 \end{aligned} \]

Hence, the derivative of \(x + a\) is \(1\).

Q3. \((px+q)\left(\dfrac{r}{x}+s\right)\)

Solution

Let the given function be \( f(x) = (px+q)\left(\dfrac{r}{x}+s\right) \). To simplify differentiation, first expand the expression algebraically.

\[ \begin{aligned} f(x) &= px \cdot \dfrac{r}{x} + psx + \dfrac{qr}{x} + qs \\ &= pr + psx + qs + \dfrac{qr}{x} \\ &= (pr+qs) + psx + \dfrac{qr}{x} \end{aligned} \]

Now differentiate each term with respect to \(x\). The terms \(p r\) and \(q s\) are constants, so their derivative is zero. The derivative of \(psx\) is \(ps\), and the derivative of \(\dfrac{qr}{x}\) is obtained using the power rule.

\[ \begin{aligned} f'(x) &= \dfrac{d}{dx}(pr+qs) + \dfrac{d}{dx}(psx) + \dfrac{d}{dx}\left(\dfrac{qr}{x}\right) \\ &= 0 + ps - \dfrac{qr}{x^2} \\ &= ps - \dfrac{qr}{x^2} \end{aligned} \]

Hence, the derivative of \((px+q)\left(\dfrac{r}{x}+s\right)\) is \( ps - \dfrac{qr}{x^2} \).

Q4. \((ax+b)(cx+d)^2\)

Solution

Let the given function be \( f(x) = (ax+b)(cx+d)^2 \). To differentiate it conveniently, first expand the expression algebraically.

\[ \begin{aligned} f(x) &= (ax+b)(cx+d)^2 \\ &= (ax+b)(c^2x^2+2cdx+d^2) \\ &= ax(c^2x^2+2cdx+d^2)+b(c^2x^2+2cdx+d^2) \\ &= ac^2x^3+2acdx^2+ad^2x+bc^2x^2+2bcdx+bd^2 \end{aligned} \]

Now differentiate term by term with respect to \(x\). The constant term disappears on differentiation.

\[ \begin{aligned} f'(x) &= \dfrac{d}{dx}\left(ac^2x^3+(2acd+bc^2)x^2+(ad^2+2bcd)x+bd^2\right) \\ &= 3ac^2x^2+2(2acd+bc^2)x+(ad^2+2bcd) \end{aligned} \]

Hence, the derivative of \((ax+b)(cx+d)^2\) is \(3ac^2x^2+2(2acd+bc^2)x+(ad^2+2bcd)\).

Q5. \(\dfrac{ax+b}{cx+d}\)

Solution

Let the given function be \( f(x) = \dfrac{ax+b}{cx+d} \). Since the function is expressed as a quotient of two differentiable functions, we apply the quotient rule to find its derivative.

The quotient rule states that if \( f(x) = \dfrac{u(x)}{v(x)} \), then \( f'(x) = \dfrac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2} \).

\[ \begin{aligned} f'(x) &= \dfrac{(cx+d)\dfrac{d}{dx}(ax+b) - (ax+b)\dfrac{d}{dx}(cx+d)}{(cx+d)^2} \\ &= \dfrac{(cx+d)a - (ax+b)c}{(cx+d)^2} \\ &= \dfrac{acx+ad-acx-bc}{(cx+d)^2} \\ &= \dfrac{ad-bc}{(cx+d)^2} \end{aligned} \]

Hence, the derivative of \(\dfrac{ax+b}{cx+d}\) is \(\dfrac{ad-bc}{(cx+d)^2}\).

Q6. \(\dfrac{1+\frac{1}{x}}{1-\frac{1}{x}}\)

Solution

Let the given function be \( f(x) = \dfrac{1+\dfrac{1}{x}}{1-\dfrac{1}{x}} \). To simplify the differentiation, first rewrite the function by multiplying the numerator and denominator by \(x\).

\[ \begin{aligned} f(x) &= \dfrac{1+\dfrac{1}{x}}{1-\dfrac{1}{x}} \\ &= \dfrac{x+1}{x-1} \end{aligned} \]

Now apply the quotient rule, since the function is expressed as a ratio of two differentiable functions.

\[ \begin{aligned} f'(x) &= \dfrac{(x-1)\dfrac{d}{dx}(x+1)-(x+1)\dfrac{d}{dx}(x-1)}{(x-1)^2} \\ &= \dfrac{(x-1)-(x+1)}{(x-1)^2} \\ &= \dfrac{x-1-x-1}{(x-1)^2} \\ &= \dfrac{-2}{(x-1)^2} \end{aligned} \]

Hence, the derivative of \( \dfrac{1+\frac{1}{x}}{1-\frac{1}{x}} \) is \( \dfrac{-2}{(x-1)^2} \).

Q7. \(\dfrac{1}{ax^2+bx+c}\)

Solution

Let the given function be defined as \( f(x)=\dfrac{1}{ax^{2}+bx+c} \). Since the function is the reciprocal of a quadratic polynomial in \(x\), its derivative can be found by applying the quotient rule.

\[ \begin{aligned} f'(x) &= \dfrac{(ax^{2}+bx+c)\dfrac{d}{dx}(1)-1\dfrac{d}{dx}(ax^{2}+bx+c)}{(ax^{2}+bx+c)^{2}} \\ &= \dfrac{0-(2ax+b)}{(ax^{2}+bx+c)^{2}} \\ &= \dfrac{-(2ax+b)}{(ax^{2}+bx+c)^{2}} \end{aligned} \]

Hence, the derivative of \( \dfrac{1}{ax^{2}+bx+c} \) is \( \dfrac{-(2ax+b)}{(ax^{2}+bx+c)^{2}} \).

Q8. \(\dfrac{ax+b}{px^2+qx+r}\)

Solution

Let the given function be defined as \( f(x)=\dfrac{ax+b}{px^{2}+qx+r} \). Since the function is expressed as a quotient of two polynomials in \(x\), its derivative is obtained by applying the quotient rule.

\[ \begin{aligned} f'(x) &= \dfrac{(px^{2}+qx+r)\dfrac{d}{dx}(ax+b)-(ax+b)\dfrac{d}{dx}(px^{2}+qx+r)}{(px^{2}+qx+r)^{2}} \\ &= \dfrac{(px^{2}+qx+r)\,a-(ax+b)(2px+q)}{(px^{2}+qx+r)^{2}} \\ &= \dfrac{apx^{2}+aqx+ar-(2apx^{2}+aqx+2bpx+bq)}{(px^{2}+qx+r)^{2}} \\ &= \dfrac{-apx^{2}-2bpx+ar-bq}{(px^{2}+qx+r)^{2}} \end{aligned} \]

Hence, the derivative of \( \dfrac{ax+b}{px^{2}+qx+r} \) is \( \dfrac{-apx^{2}-2bpx+ar-bq}{(px^{2}+qx+r)^{2}} \).

Q9. \(\dfrac{px^2+qx+r}{ax+b}\)

Solution

Let the given function be defined as \( f(x)=\dfrac{px^{2}+qx+r}{ax+b} \). Since the function is a quotient of two polynomials in \(x\), its derivative is obtained by applying the quotient rule.

\[ \begin{aligned} f'(x) &= \dfrac{(ax+b)\dfrac{d}{dx}(px^{2}+qx+r)-(px^{2}+qx+r)\dfrac{d}{dx}(ax+b)}{(ax+b)^{2}} \\ &= \dfrac{(ax+b)(2px+q)-(px^{2}+qx+r)a}{(ax+b)^{2}} \\ &= \dfrac{2apx^{2}+aqx+2bpx+bq-apx^{2}-aqx-ar}{(ax+b)^{2}} \\ &= \dfrac{apx^{2}+2bpx+bq-ar}{(ax+b)^{2}} \end{aligned} \]

Hence, the derivative of \( \dfrac{px^{2}+qx+r}{ax+b} \) is \( \dfrac{apx^{2}+2bpx+bq-ar}{(ax+b)^{2}} \).

Q10. \(\dfrac{a}{x^4}-\dfrac{b}{x^2}+\cos x\)

Solution

Let the given function be defined as \( f(x)=\dfrac{a}{x^{4}}-\dfrac{b}{x^{2}}+\cos x \). Since the function is a sum of functions of \(x\), its derivative can be found by differentiating each term separately.

\[ \begin{aligned} f'(x) &= \dfrac{d}{dx}(ax^{-4})-\dfrac{d}{dx}(bx^{-2})+\dfrac{d}{dx}(\cos x) \\ &= -4ax^{-5}+2bx^{-3}-\sin x \\ &= \dfrac{-4a}{x^{5}}+\dfrac{2b}{x^{3}}-\sin x \end{aligned} \]

Hence, the derivative of \( \dfrac{a}{x^{4}}-\dfrac{b}{x^{2}}+\cos x \) is \( \dfrac{-4a}{x^{5}}+\dfrac{2b}{x^{3}}-\sin x \).

Q11. \(4\sqrt{x}-2\)

Solution

Let the given function be defined as \( f(x)=4\sqrt{x}-2 \). This function consists of a power of \(x\) and a constant, so its derivative can be obtained by applying the power rule and the fact that the derivative of a constant is zero.

\[ \begin{aligned} f'(x) &= \dfrac{d}{dx}(4x^{1/2})-\dfrac{d}{dx}(2) \\ &= 4\cdot \dfrac{1}{2}x^{1/2-1}-0 \\ &= 2x^{-1/2} \\ &= \dfrac{2}{\sqrt{x}} \end{aligned} \]

Hence, the derivative of \( 4\sqrt{x}-2 \) is \( \dfrac{2}{\sqrt{x}} \).

Q12. \((ax+b)^n\)

Solution

Let the given function be defined as \( f(x)=(ax+b)^{n} \), where \(a\), \(b\), and \(n\) are constants. Since the function is a power of a linear expression in \(x\), its derivative is obtained by applying the chain rule.

\[ \begin{aligned} f'(x) &= \dfrac{d}{dx}(ax+b)^{n} \\ &= n(ax+b)^{\,n-1}\cdot \dfrac{d}{dx}(ax+b) \\ &= n(ax+b)^{\,n-1}\cdot a \\ &= an(ax+b)^{\,n-1} \end{aligned} \]

Hence, the derivative of \( (ax+b)^{n} \) is \( an(ax+b)^{\,n-1} \).

Q13. \((ax+b)^n(cx+d)^m\)

Solution

Let the given function be defined as \( f(x)=(ax+b)^{n}(cx+d)^{m} \). Since the function is a product of two functions of \(x\), its derivative is obtained by applying the product rule.

\[ \begin{aligned} f'(x) &= (cx+d)^{m}\dfrac{d}{dx}(ax+b)^{n}+(ax+b)^{n}\dfrac{d}{dx}(cx+d)^{m} \\ &= (cx+d)^{m}\cdot na(ax+b)^{n-1}+(ax+b)^{n}\cdot mc(cx+d)^{m-1} \\ &= (ax+b)^{n-1}(cx+d)^{m-1}\left[na(cx+d)+mc(ax+b)\right] \end{aligned} \]

Hence, the derivative of \( (ax+b)^{n}(cx+d)^{m} \) is \( (ax+b)^{n-1}(cx+d)^{m-1}\left[na(cx+d)+mc(ax+b)\right] \).

Q14. \(\sin(x+a)\)

Solution

Let the given function be defined as \( f(x)=\sin(x+a) \), where \(a\) is a constant. Since the argument of the sine function is a linear function of \(x\), the derivative is obtained by applying the chain rule.

\[ \begin{aligned} f'(x) &= \dfrac{d}{dx}\sin(x+a) \\ &= \cos(x+a)\cdot \dfrac{d}{dx}(x+a) \\ &= \cos(x+a) \end{aligned} \]

Hence, the derivative of \( \sin(x+a) \) is \( \cos(x+a) \).

Q15. \(\text{cosec }x\cot x\)

Solution

Let the given function be defined as \( f(x)=\text{cosec } x\cot x \). Since the function is a product of two trigonometric functions of \(x\), its derivative is obtained by applying the product rule.

\[ \begin{aligned} f'(x) &= \cot x\dfrac{d}{dx}(\text{cosec } x)+\text{cosec } x\dfrac{d}{dx}(\cot x) \\ &= \cot x(-\text{cosec } x\cot x)+\text{cosec } x(-\text{cosec }^{2}x) \\ &= -\text{cosec } x\cot^{2}x-\text{cosec }^{3}x \end{aligned} \]

Hence, the derivative of \( \text{cosec } x\cot x \) is \( -\text{cosec }^{3}x-\text{cosec } x\cot^{2}x \).

Q16. \(\dfrac{\cos x}{a+\sin x}\)

Solution

Let the given function be defined as \( f(x)=\dfrac{\cos x}{a+\sin x} \), where \(a\) is a constant. Since the function is a quotient of two functions of \(x\), its derivative is obtained by applying the quotient rule.

\[ \begin{aligned} f'(x) &= \dfrac{(a+\sin x)\dfrac{d}{dx}(\cos x)-\cos x\dfrac{d}{dx}(a+\sin x)}{(a+\sin x)^{2}} \\ &= \dfrac{(a+\sin x)(-\sin x)-\cos x\cdot \cos x}{(a+\sin x)^{2}} \\ &= \dfrac{-a\sin x-\sin^{2}x-\cos^{2}x}{(a+\sin x)^{2}} \\ &= \dfrac{-a\sin x-1}{(a+\sin x)^{2}} \end{aligned} \]

Thus, the derivative of \( \dfrac{\cos x}{a+\sin x} \) is \( \dfrac{-a\sin x-1}{(a+\sin x)^{2}} \).

Q17. \(\dfrac{\sin x+\cos x}{\sin x - \cos x}\)

Solution

Let the given function be defined as \( f(x)=\dfrac{\sin x+\cos x}{\sin x-\cos x} \). Since the function is a quotient of two trigonometric expressions, its derivative is obtained by applying the quotient rule.

\[ \begin{aligned} f'(x) &= \dfrac{(\sin x-\cos x)\dfrac{d}{dx}(\sin x+\cos x)-(\sin x+\cos x)\dfrac{d}{dx}(\sin x-\cos x)}{(\sin x-\cos x)^{2}} \\ &= \dfrac{(\sin x-\cos x)(\cos x-\sin x)-(\sin x+\cos x)(\cos x+\sin x)}{(\sin x-\cos x)^{2}} \\ &= \dfrac{-(\sin x-\cos x)^{2}-(\sin x+\cos x)^{2}}{(\sin x-\cos x)^{2}} \\ &= \dfrac{-\left(\sin^{2}x+\cos^{2}x-2\sin x\cos x\right)-\left(\sin^{2}x+\cos^{2}x+2\sin x\cos x\right)}{(\sin x-\cos x)^{2}} \\ &= \dfrac{-2}{(\sin x-\cos x)^{2}} \end{aligned} \]

Hence, the derivative of \( \dfrac{\sin x+\cos x}{\sin x-\cos x} \) is \( \dfrac{-2}{(\sin x-\cos x)^{2}} \).

Q18. \(\dfrac{\sec x -1}{\sec x +1}\)

Solution

Let the given function be defined as \( f(x)=\dfrac{\sec x-1}{\sec x+1} \). Since the function is a quotient of two functions of \(x\), its derivative is obtained by applying the quotient rule.

\[ \begin{aligned} f'(x) &= \dfrac{(\sec x+1)\dfrac{d}{dx}(\sec x-1)-(\sec x-1)\dfrac{d}{dx}(\sec x+1)}{(\sec x+1)^{2}} \\ &= \dfrac{(\sec x+1)(\sec x\tan x)-(\sec x-1)(\sec x\tan x)}{(\sec x+1)^{2}} \\ &= \dfrac{\sec x\tan x\left[(\sec x+1)-(\sec x-1)\right]}{(\sec x+1)^{2}} \\ &= \dfrac{2\sec x\tan x}{(\sec x+1)^{2}} \end{aligned} \]

Hence, the derivative of \( \dfrac{\sec x-1}{\sec x+1} \) is \( \dfrac{2\sec x\tan x}{(\sec x+1)^{2}} \).

Q19. \(\sin^n x\)

Solution

Let the given function be defined as \( f(x)=\sin^{n}x \), where \(n\) is a constant. This is a composite function in which \(\sin x\) is raised to the power \(n\), so its derivative is obtained by applying the chain rule.

\[ \begin{aligned} f'(x) &= \dfrac{d}{dx}(\sin x)^{n} \\ &= n(\sin x)^{\,n-1}\cdot \dfrac{d}{dx}(\sin x) \\ &= n\sin^{\,n-1}x\cdot \cos x \end{aligned} \]

Hence, the derivative of \( \sin^{n}x \) is \( n\sin^{\,n-1}x\cos x \).

Q20. \(\dfrac{a+b\sin x}{c+d\cos x}\)

Solution

Let the given function be defined as \( f(x)=\dfrac{a+b\sin x}{c+d\cos x} \). Since the function is a quotient of two differentiable functions of \(x\), its derivative is obtained by applying the quotient rule.

\[ \begin{aligned} f'(x) &= \dfrac{(c+d\cos x)\dfrac{d}{dx}(a+b\sin x)-(a+b\sin x)\dfrac{d}{dx}(c+d\cos x)}{(c+d\cos x)^{2}} \\ &= \dfrac{(c+d\cos x)(b\cos x)-(a+b\sin x)(-d\sin x)}{(c+d\cos x)^{2}} \\ &= \dfrac{bc\cos x+bd\cos^{2}x+ad\sin x+bd\sin^{2}x}{(c+d\cos x)^{2}} \\ &= \dfrac{bc\cos x+ad\sin x+bd(\cos^{2}x+\sin^{2}x)}{(c+d\cos x)^{2}} \\ &= \dfrac{bc\cos x+ad\sin x+bd}{(c+d\cos x)^{2}} \end{aligned} \]

Hence, the derivative of \( \dfrac{a+b\sin x}{c+d\cos x} \) is \( \dfrac{bc\cos x+ad\sin x+bd}{(c+d\cos x)^{2}} \).

Q21. \(\dfrac{\sin (x+a)}{\cos a}\)

Solution

Let the given function be defined as \( f(x)=\dfrac{\sin(x+a)}{\cos a} \), where \(a\) is a constant. Since \(\cos a\) is a constant, the function is a constant multiple of \(\sin(x+a)\), and its derivative can be found directly.

\[ \begin{aligned} f'(x) &= \dfrac{1}{\cos a}\dfrac{d}{dx}\sin(x+a) \\ &= \dfrac{1}{\cos a}\cdot \cos(x+a) \\ &= \dfrac{\cos(x+a)}{\cos a} \end{aligned} \]

Hence, the derivative of \( \dfrac{\sin(x+a)}{\cos a} \) is \( \dfrac{\cos(x+a)}{\cos a} \).

Q22. \(x^4(5\sin x-3\cos x)\)

Solution

Let the given function be defined as \( f(x)=x^{4}(5\sin x-3\cos x) \). First, expand the expression to make differentiation clearer.

\[ \begin{aligned} f(x) &= 5x^{4}\sin x-3x^{4}\cos x \end{aligned} \]

Differentiate each term separately using the product rule, since each term is a product of a polynomial and a trigonometric function.

\[ \begin{aligned} f'(x) &= \dfrac{d}{dx}(5x^{4}\sin x)-\dfrac{d}{dx}(3x^{4}\cos x) \end{aligned} \]

\[ \begin{aligned} \dfrac{d}{dx}(5x^{4}\sin x) &= 5\dfrac{d}{dx}(x^{4}\sin x) \\ &= 5\left[\sin x\dfrac{d}{dx}(x^{4})+x^{4}\dfrac{d}{dx}(\sin x)\right] \\ &= 5\left[4x^{3}\sin x+x^{4}\cos x\right] \\ &= 20x^{3}\sin x+5x^{4}\cos x \end{aligned} \]

\[ \begin{aligned} \dfrac{d}{dx}(3x^{4}\cos x) &= 3\left[\cos x\dfrac{d}{dx}(x^{4})+x^{4}\dfrac{d}{dx}(\cos x)\right] \\ &= 3\left[4x^{3}\cos x-x^{4}\sin x\right] \\ &= 12x^{3}\cos x-3x^{4}\sin x \end{aligned} \]

\[ \begin{aligned} f'(x) &= 20x^{3}\sin x+5x^{4}\cos x-12x^{3}\cos x+3x^{4}\sin x \\ &= x^{3}(20\sin x-12\cos x)+x^{4}(5\cos x+3\sin x) \end{aligned} \]

Hence, the derivative of \( x^{4}(5\sin x-3\cos x) \) is \( x^{3}(20\sin x-12\cos x)+x^{4}(5\cos x+3\sin x) \).

Q23. \(\left(x^2+1\right)\cos x\)

Solution

Let the given function be defined as \( f(x)=(x^{2}+1)\cos x \). Since the function is a product of a polynomial and a trigonometric function, its derivative is obtained by applying the product rule.

\[ \begin{aligned} f'(x) &= \cos x\dfrac{d}{dx}(x^{2}+1)+(x^{2}+1)\dfrac{d}{dx}(\cos x) \\ &= \cos x\cdot 2x+(x^{2}+1)(-\sin x) \\ &= 2x\cos x-(x^{2}+1)\sin x \\ &= -x^{2}\sin x-\sin x+2x\cos x \end{aligned} \]

Hence, the derivative of \( (x^{2}+1)\cos x \) is \( -x^{2}\sin x-\sin x+2x\cos x \).

Q24. \(\left(ax^2+\sin x\right)\left(p+q\cos x\right)\)

Solution

Let the given function be defined as \( f(x)=(ax^{2}+\sin x)(p+q\cos x) \). Since the function is a product of two functions of \(x\), its derivative is obtained by applying the product rule.

\[ \begin{aligned} f'(x) &= (p+q\cos x)\dfrac{d}{dx}(ax^{2}+\sin x)+(ax^{2}+\sin x)\dfrac{d}{dx}(p+q\cos x) \\ &= (p+q\cos x)(2ax+\cos x)-(ax^{2}+\sin x)(q\sin x) \end{aligned} \]

Hence, the derivative of \( (ax^{2}+\sin x)(p+q\cos x) \) is \( (p+q\cos x)(2ax+\cos x)-(ax^{2}+\sin x)(q\sin x) \).

Q25. \((x+\cos x)(x-\tan x)\)

Solution

Let the given function be defined as \( f(x)=(x+\cos x)(x-\tan x) \). Since the function is a product of two functions of \(x\), its derivative is obtained by applying the product rule.

\[ \begin{aligned} f'(x) &= (x-\tan x)\dfrac{d}{dx}(x+\cos x)+(x+\cos x)\dfrac{d}{dx}(x-\tan x) \\ &= (x-\tan x)(1-\sin x)+(x+\cos x)(1-\sec^{2}x) \end{aligned} \]

\[ \begin{aligned} f'(x) &= (x-\tan x)(1-\sin x)-\tan^{2}x(x+\cos x) \end{aligned} \]

Hence, the derivative of \( (x+\cos x)(x-\tan x) \) is \( (x-\tan x)(1-\sin x)-\tan^{2}x(x+\cos x) \).

Q26. \(\dfrac{4x+5\sin x}{3x+7\cos x}\)

Solution

Let the given function be defined as \( f(x)=\dfrac{4x+5\sin x}{3x+7\cos x} \). Since the function is a quotient of two differentiable functions of \(x\), its derivative is obtained by applying the quotient rule.

\[ \begin{aligned} f'(x) &= \dfrac{(3x+7\cos x)\dfrac{d}{dx}(4x+5\sin x)-(4x+5\sin x)\dfrac{d}{dx}(3x+7\cos x)}{(3x+7\cos x)^{2}} \\ &= \dfrac{(3x+7\cos x)(4+5\cos x)-(4x+5\sin x)(3-7\sin x)}{(3x+7\cos x)^{2}} \\ &= \dfrac{12x+15x\cos x+28\cos x+35\cos^{2}x-12x+28x\sin x-15\sin x+35\sin^{2}x}{(3x+7\cos x)^{2}} \\ &= \dfrac{15x\cos x+28x\sin x+28\cos x-15\sin x+35}{(3x+7\cos x)^{2}} \end{aligned} \]

Hence, the derivative of \( \dfrac{4x+5\sin x}{3x+7\cos x} \) is \( \dfrac{15x\cos x+28x\sin x+28\cos x-15\sin x+35}{(3x+7\cos x)^{2}} \).

Q27. \(\dfrac{x^2\cos \left(\dfrac{\pi}{4}\right)}{\sin x}\)

Solution

Let the given function be defined as \( f(x)=\dfrac{x^{2}\cos \left(\dfrac{\pi}{4}\right)}{\sin x} \). Since the function is a quotient of two functions of \(x\), its derivative is obtained by applying the quotient rule. Note that \( \cos\left(\dfrac{\pi}{4}\right) \) is a constant.

\[ \begin{aligned} f'(x) &= \dfrac{\sin x\,\dfrac{d}{dx}\left(x^{2}\cos \dfrac{\pi}{4}\right)-x^{2}\cos \dfrac{\pi}{4}\,\dfrac{d}{dx}(\sin x)}{\sin^{2}x} \end{aligned} \]

\[ \begin{aligned} \dfrac{d}{dx}\left(x^{2}\cos \dfrac{\pi}{4}\right) &= \cos \dfrac{\pi}{4}\cdot \dfrac{d}{dx}(x^{2}) \\ &= 2x\cos \dfrac{\pi}{4} \end{aligned} \]

\[ \begin{aligned} f'(x) &= \dfrac{\sin x\left(2x\cos \dfrac{\pi}{4}\right)-x^{2}\cos \dfrac{\pi}{4}\cos x}{\sin^{2}x} \\ &= \dfrac{\cos \dfrac{\pi}{4}\left(2x\sin x-x^{2}\cos x\right)}{\sin^{2}x} \end{aligned} \]

Hence, the derivative of \( \dfrac{x^{2}\cos \left(\dfrac{\pi}{4}\right)}{\sin x} \) is \( \dfrac{\cos \left(\dfrac{\pi}{4}\right)\left(2x\sin x-x^{2}\cos x\right)}{\sin^{2}x} \).

Q28. \(\dfrac{x}{1+\tan x}\)

Solution

Let the given function be defined as \( f(x)=\dfrac{x}{1+\tan x} \). Since the function is expressed as a quotient of two functions of \(x\), its derivative is obtained by applying the quotient rule.

\[ \begin{aligned} f'(x) &= \dfrac{(1+\tan x)\dfrac{d}{dx}(x)-x\dfrac{d}{dx}(1+\tan x)}{(1+\tan x)^{2}} \\ &= \dfrac{(1+\tan x)-x\sec^{2}x}{(1+\tan x)^{2}} \\ &= \dfrac{1+\tan x-x\sec^{2}x}{(1+\tan x)^{2}} \end{aligned} \]

Hence, the derivative of \( \dfrac{x}{1+\tan x} \) is \( \dfrac{1+\tan x-x\sec^{2}x}{(1+\tan x)^{2}} \).

Q29. \((x+\sec x)(x-\tan x)\)

Solution

Let the given function be defined as \( f(x)=(x+\sec x)(x-\tan x) \). Since the function is expressed as a product of two differentiable functions of \(x\), its derivative is obtained by applying the product rule.

\[ \begin{aligned} f'(x) &= (x-\tan x)\dfrac{d}{dx}(x+\sec x)+(x+\sec x)\dfrac{d}{dx}(x-\tan x) \\ &= (x-\tan x)\left(1+\sec x\tan x\right)+(x+\sec x)\left(1-\sec^{2}x\right) \end{aligned} \]

This expression gives the required derivative of the function \( (x+\sec x)(x-\tan x) \).

Q30. \(\dfrac{x}{\sin^n x}\)

Solution

Let the given function be defined as \( f(x)=\dfrac{x}{\sin^{n}x} \). This function can be treated as a quotient of two differentiable functions of \(x\), so its derivative is obtained by applying the quotient rule.

\[ \begin{aligned} f'(x) &= \dfrac{\sin^{n}x\,\dfrac{d}{dx}(x)-x\,\dfrac{d}{dx}(\sin^{n}x)}{\sin^{2n}x} \\ &= \dfrac{\sin^{n}x-x\cdot n\sin^{n-1}x\cos x}{\sin^{2n}x} \\ &= \dfrac{\sin^{n-1}x\left(\sin x-nx\cos x\right)}{\sin^{2n}x} \\ &= \dfrac{\sin x-nx\cos x}{\sin^{n+1}x} \end{aligned} \]

Hence, the derivative of \( \dfrac{x}{\sin^{n}x} \) is \( \dfrac{\sin x-nx\cos x}{\sin^{n+1}x} \).