H.S.C.
Differentiation EX 1.2 Q.1 Maharashtra state Board Mathematics and statistic for arts and science
Chapter 1 - Differentiation
Exercise 1.2 Q.1
$\begin{array}{rl}
y&=\sqrt{x} \\
y&=x^{\frac{1}{2}}\\
&Squaring \quad both \quad sides \\
y^{2}&=x\\
x&=y^{2}\\
&Diff \quad w.r.t. \quad y\\
\frac{d x}{d y}&=\frac{d}{d y} \cdot\left(y^{2}\right)\\
\frac{d x}{d y}&=2 y\\
Now, \\
\frac{d y}{d x}&=\frac{1}{\left(\frac{d x}{d y}\right)}\\
\frac{d y}{d x}&=\frac{1}{2 y}\end{array}$
$\begin{array}{rl} y&=\sqrt{2-\sqrt{x}}\\ &Squaring \quad both \quad side \\ y^{2}&=2-\sqrt{x}\\ \sqrt{x}&=2-y^{2}\\ &Squaring \quad both \quad side \\ x&=\left(2-y^{2}\right)^{2}=f^{-1}(y)\\ &diff \quad w.r.t \quad y \\ \frac{d x}{d y}&=\frac{d}{d y}\left(2-y^{2}\right)^{2}\\ &=2\left(2-y^{2}\right) \cdot(-2 y)\\ \frac{d x}{d y}&=-4 y\left(2-y^{2}\right)\\\\ \therefore \frac{d y}{d x}&=\frac{1}{\left(\frac{d x}{d y}\right)}\\ &=\frac{1}{-4 \sqrt{2-\sqrt{x}} \cdot \sqrt{x}}\\ &=\frac{1}{-4 \sqrt{x} \sqrt{2-\sqrt{x}}}\end{array}$
$\begin{array}{rl} y&=\sqrt[3]{x-2} \\ &cubing \quad both \quad side \quad \\ y^{3}&=x-2\\ x&=y^{3}+2=f^{-1}(y)\\ &Diff \quad w.r.t \quad y \\ \frac{d x}{d y}&=\frac{d}{d y}\left(y^{3}+2\right)\\ \frac{d x}{d y}&=3 \cdot y^{2}\\ \frac{d y}{d x}&=\frac{1}{\left(\frac{d x}{d y}\right)}\\ &=\frac{1}{3 y^{2}}\\ &=\frac{1}{3[\sqrt[3]{x-2}]^{2}}\\ &=\frac{1}{3 \sqrt[3]{(x-2)^{2}}}\end{array}$
$\begin{array}{rl} y&=\log (2 x-1)\\ &Taking \quad antilog \quad on \quad both \quad side\\ e^{y}&=2 x-1\\ 2 x&=e^{y}+1\\ x&=\frac{e^{y}+1}{2}=f^{-1}(y)\\ &Diff \quad w.r.t \quad y \\ \frac{d x}{d y}&=\frac{1}{2} \frac{d}{d y}\left(e^{y}+1\right)\\ \frac{d x}{d y}&=\frac{1}{2} \cdot e^{y}\\\\ \frac{d y}{d x}&=\frac{1}{\left(\frac{d x}{d y}\right)}\\ &=\frac{1}{\frac{1}{2}\left(e^{y}\right)}\\ \frac{d y}{d x}&=\frac{2}{e^{y}}\\ &=\frac{2}{2 x-1}\end{array}$
$\begin{aligned}
y&=2 x+3\\
2 x &=y-3 \\
x &=\frac{y-3}{2}=f^{-1}(y) \\
&\text { Diff }\text { w.r.t }\quad y \\
\frac{d x}{d y} &=\frac{1}{2} \cdot \frac{d}{d y}(y-3) \\
\frac{d x}{d y} &=\frac{1}{2}(1) \\
\frac{d x}{d y} &=\frac{1}{2} \\
\text { Now }, \\
\frac{d y}{d x} &=\frac{1}{\left(\frac{d x}{d y}\right)} \\ &=\frac{1}{(1 / 2)} \\ \frac{d y}{d x} &=2 \end{aligned}$
$\begin{aligned} y&=e^{x}-3\\ \quad e^{x}&=y+3\\ &Taking \quad log \quad on \quad both \quad Side \\ \log \left(e^{x}\right) &=\log (y+3) \\ x &=\log (y+3)=f^{-1}(y) \\ &Diff \quad w.r.t \quad y \\ \frac{d x}{d y} &=\frac{d}{d y} \log (y+3) \\ \frac{d x}{d y} &=\frac{1}{y+3} \\\\ NOW,\\ \frac{d y}{d x}&=\frac{1}{\left(\frac{d x}{d y}\right)}\\ &=\frac{1}{\left(\frac{1}{y+3}\right)}\\ &=y+3\\ &=e^{x}-3+3\\ &=e^{x} \end{aligned}$
$\begin{aligned} y&=e^{2 x-3} \\ &Taking \quad log \quad on \quad both \\ \log y&=\log e^{(2 x-3)}\\ \log y&=2 x-3\\ 2 x&=\log y+3\\ x&=\frac{\log y+3}{2}\\ &Diff \quad w.r.t \quad y \\ \frac{d x}{d y}&=\frac{1}{2} \cdot \frac{d}{d y}(\log y+3)\\ &=\frac{1}{2} \times \frac{1}{y}\\ \frac{d x}{d y}&=\frac{1}{2 y}\\ Now, \\ \frac{d y}{d x}&=\frac{1}{\left(\frac{d x}{d y}\right)}\\ &=\frac{1}{\left(\frac{1}{24}\right)}\\ &=2 y\\ &=2 \cdot\left(e^{2 x-3}\right)\end{aligned}$
$\begin{aligned} y&=\log _{2}\left(\frac{x}{2}\right)\\ 2^{y} &=\frac{x}{2} \\ x &=2 \cdot 2^{y} \\ x &=2^{1+y}=f^{-1}(y) \\ &Diff \quad w.r.t. \quad y \\ \frac{d x}{d y} &=\frac{d}{d y}\left(2^{(1+y)}\right) \\ &Using=a^{x} \log a \\ \frac{d x}{d y} &={2^{1+y}} \log 2 \\ \frac{d x}{d y} &=\frac{1}{\left(\frac{d x}{d y}\right)} \\ &=\frac{1}{\left(2^{1+y}\right) \log 2} \\ &=\frac{1}{x \log 2}\end{aligned}$
Exercise 1.2
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