Chapter 1 - Differentiation
Exercise 1.2
Q.3
i)
$\begin{aligned}
y&=x^{5}+2 x^{3}+3 x\\
& Diff \quad w.r.t. \quad x \\
\frac{d y}{d x}&=\frac{d}{d x}\left(x^{5}+2 x^{3}+3 x\right)\\
\frac{d y}{d x}&=5 x^{4}+6 x^{2}+3\\\\
Now,\\
\frac{d x}{d y}&=\frac{1}{\left(\frac{d y}{d x}\right)}\\
&=\frac{1}{5 x^{4}+6 x^{2}+3}\\\\
at ,x=1 \\
y&=(1)^{5}+2(1)^{3}+3(1)\\
y&=1+2+3\\
y&=6 & \\\\
\therefore
\left[\frac{d x}{d y}\right]_{y=6} &=\frac{1}{\left[\frac{d y}{d x}\right]_{x=1}} \\ &=\frac{1}{5(1)^{4}+6(1)^{2}+3} \\ &=\frac{1}{5+6+3} \\ &=\frac{1}{14} \end{aligned}$
ii)
$\begin{aligned}
y&=e^{x}+3 x+2\\
&Diff \quad w.r.t. \quad x \\
\frac{d y}{d x}&=\frac{d}{d x}\left(e^{x}+3 x+2\right)\\
\frac{d y}{d x}&=e^{x}+3\\\\
Now, \\
\frac{d x}{d y}&=\frac{1}{\left(\frac{d y}{d x}\right)}\\
&=\frac{1}{e^{x}+3}\\
at,x=0\\
y&=e^{0}+3(0)+2\\
&=1+2\\
&=3\\\\
\left[\frac{d x}{d y}\right]_{y=3} &=\frac{1}{\left(\frac{d y}{d x}\right)_{x=0}}\\
&=\frac{1}{e^{0}+3} \\ &=\frac{1}{1+3} \\ &=\frac{1}{4} \end{aligned}$
iii)
$\begin{aligned}
y&=3 x^{2}+2 \log x^{3}\\
y&=3 x^{2}+2 \times 3 \log x\\
&Diff \quad w.r.t. \quad x \\
\frac{d y}{d x} &=6 x+6 \frac{1}{x} \\
&=6 x+\frac{6}{x} \\
&=\frac{6 x^{2}+6}{x} \\
Now,\\
\frac{d x}{d y}&=\frac{1}{\left(\frac{d y}{d x}\right)}\\
&=\frac{1}{\left(\frac{6 x^{2}+6}{x}\right)}\\
&=\frac{x}{6\left(x^{2}+1\right)}\\
at, x=1\\
y&=3(1)^{2}+2 \cdot \log 1\\
y &=3+2 \cdot \log 1 \\
y&=3 \\
\\\left[\frac{d y}{d y}\right]_{y=3} &=\frac{1}{\left[\frac{d y}{d x}\right]_{x=1}} \\ &=\frac{1}{6(1+1)} \\ &=\frac{1}{12} \end{aligned}$
iV)
$\begin{aligned}
y&=\sin (x-2)+x^{2}\\
&Diff \quad w.r.t. \quad x \\
\frac{d y}{d x}&=\frac{d}{d x}\left[\sin (x-2)+x^{2}\right]\\
\frac{d y}{d x}&=\cos (x-2)+2 x\\\\
Now, \\
\frac{d x}{d y}&=\frac{1}{\left(\frac{d y}{d x}\right)}\\
&=\frac{1}{\cos (x-2)+2 x}\\
at,x=2\\
y&=\sin (2-2)+2^{2}\\
y&=\sin 0+4\\
y&=4\\\\
\left[\frac{d x}{d y}\right]_{y=4} &=\frac{1}{\left[\frac{d y}{d x}\right]_{x=2}} \\ \\
&=\frac{1}{\cos (2-2)+2 \cdot 2} \\ &=\frac{1}{\cos 0+4} \\ &=\frac{1}{1+4} \\ &=\frac{1}{5} \end{aligned}$
Exercise 1.2
