Product, Quotient & Chain Rule – Practice ( Qn 6 – 10)

Question No: 6

Find the differentiation of \(y=e^{x^2-4 x+11}\)
Answer : \( (2 x-4) e^{x^2-4 x+11}\)
. \( \begin{aligned} y & =e^{x^2-4 x+11} \\ \frac{d y}{d x} & =e^{x^2-4 x+11} \times \frac{d}{d x}\left(x^2-4 x+11\right) \\ & =e^{x^2-4 x+11} \times(2 x-4) \\ & =(2 x-4) e^{x^2-4 x+11} \end{aligned}\) .

.

.

Question No: 7

Find the differentiation of \(e^{3 x} \cdot \ln 4 x\)
Answer : \( e^{3 x}\left[\frac{1}{x}+3 \ln 4 x\right]\)
. \( \begin{aligned} y & =e^{3 x} \cdot \ln 4 x \\ \frac{d y}{d x} & =e^{3 x} \times \frac{1}{A x} \times 4+\ln 4 x \times e^{3 x} \times 3 \\ & =\frac{e^{3 x}}{x}+3 e^{3 x} \ln 4 x \\ \frac{d y}{d x} & =e^{3 x}\left[\frac{1}{x}+3 \ln 4 x\right] \end{aligned}\) .

.

.

Question No: 8

Use the quotient rule to find the differentiation of \(f(x)=\frac{2 \ln x+1}{x-3}\)
Answer : \( \frac{x-6-2 x \ln x}{x(x-3)^2}\)
. \( \begin{aligned} y & =\frac{2 \ln x+1}{x-3} \\ \frac{d y}{d x} & =\frac{(x-3) \frac{2}{x}-(2 \ln x+1)(1)}{(x-3)^2} \\ & =\frac{2 x-6-2 x \ln x-x}{(x-3)^2} \\ & =\frac{x-6-2 x \ln x}{x(x-3)^2} \end{aligned}\) .

.

.

Question No: 9

Use the derivatives of \( \sin x\) and \( \cos x\) to find that the derivative of \( \tan x\)
Answer : \( \frac{1}{\cos ^2 x}\)
. \( \begin{aligned} \frac{d}{d x}(\tan x) & =\frac{d}{d x}\left(\frac{\sin x}{\cos x}\right) \\ & =\frac{\cos x \cdot \cos x-\sin x \times-\sin x}{\cos ^2 x} \\ & =\frac{\cos ^2 x+\sin ^2 x}{\cos ^2 x} \\ & =\frac{1}{\cos ^2 x} \\ & =\sec ^2 x \end{aligned}\) .

.

.

Question No: 10

Find the differentiation of \(y=\left(x^2-3 x+5\right)^7 \)
Answer : \( (14 x-21)\left(x^2-3 x+5\right)^6\)
. \( \begin{aligned} y & =\left(x^2-3 x+5\right)^7 \\ \frac{d y}{d x} & =\mathbb{R}\left(x^2-3 x+5\right)^6 \times(2 x-5) \\ & =(14 x-21)\left(x^2-3 x+5\right)^6 \end{aligned}\) .