Chapter 1 - Differentiation
Exercise 1.3
Q.3 Differentiate the following w. r. t. x
i)
$\begin{aligned}
\sqrt{x}+\sqrt{y}=\sqrt{a} \\
Sol^n\\
Diff \quad w.r.t \quad x \\
\frac{d}{d x} \sqrt x+\frac{d}{d x} \sqrt{y}& =0 \\
\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \frac{d y}{d x}&=0\\
\frac{1}{2 \sqrt{y}} \frac{d y}{d x}=-\frac{1}{2\sqrt{x}} \\
\frac{d y}{d x}=-\sqrt\frac{y}{x}
\end{aligned}$
ii)
$\begin{aligned}
&x \sqrt{x}+y\sqrt{y}=a \sqrt{a}\\
&x \cdot x^{\frac{1}{2}}+y \cdot y^{\frac{1}{2}}=a \cdot a^{\frac{1}{2}}\\
&x^{\frac{3}{2}}+y^{\frac{3}{2}}=a^{\frac{3}{2}}\\
&Diff \quad w.r.t \quad x\\
&\frac{3}{2} x^{\frac{3}{2}-1}+\frac{3}{2} \cdot y^{\frac{3}{2}-1} \cdot \frac{d y}{d x}=\frac{d}{d x} a^{3 / 2}\\
&\frac{3}{2} x^{\frac{1}{2}}+\frac{3}{2} y^{\frac{1}{2}} \frac{d y}{d x}=0\\
&\frac{3}{2} \sqrt{y} \frac{d y}{d x} =-\frac{3}{2} \sqrt{x} \\ &\sqrt{y} \frac{d y}{d x}=-\sqrt{x} \\
&\frac{d y}{d x} =-\sqrt{\frac{x}{y}} \end{aligned}$
iii)
$\begin{aligned}
&x+\sqrt{x y}+y=1\\Sol^n\\
&Diff \quad wrt \quad x\\
&\frac{d}{d x}(x)+\frac{d}{d x} \sqrt{x y}+\frac{d}{d x} y=\frac{d}{d x}(1)\\
&1+\frac{1}{2 \sqrt{x y}} \frac{d}{d x}(x \cdot y)+\frac{d y}{d x}=0\\
&1+\frac{1}{2 \sqrt{x y}}\left[x \cdot \frac{d y}{d x}+y \cdot \frac{d}{d x}(x)\right]+\frac{d y}{d x}=0\\
&1+\frac{1}{2 \sqrt{x y}}\left[x \cdot \frac{d y}{d x}+y\right]+\frac{d y}{d x}=0\\
&\frac{d y}{d x}\left[\frac{x}{2 \sqrt{x y}}+1\right]+1+\frac{y}{2 \sqrt{x y}}=0\\
&\frac{d y}{d x}\left[\frac{1}{2} {\frac{\sqrt x}{\sqrt y}}+1\right]+1+\frac{1}{2} {\frac{\sqrt y}{\sqrt x}}=0\\
&\frac{d y}{d x}\left[\frac{\sqrt{x}+2\sqrt{y}}{2 \sqrt{y}}\right] =-1-\frac{\sqrt{y}}{2 \sqrt{x}} \\ \\
&\frac{d y}{d x}\left[\frac{\sqrt{x}+2\sqrt{y}}{2 \sqrt{y}}\right] =-\frac{2 \sqrt{x}+{\sqrt{y}}}{2 \sqrt{x}} \\ \\
&\frac{d y}{d x}=-\frac{\frac{2 \sqrt{x}+\sqrt{y}}{2 \sqrt{x}}}{\frac{2 \sqrt{y}+\sqrt{x}}{2 \sqrt{y}}} \\ \\
&\frac{d y}{d x}=-\frac{\sqrt{y}(2 \sqrt{x}+\sqrt{y})}{\sqrt{x}(2 \sqrt{y}+\sqrt{x})} \end{aligned}$
iv)
$\begin{aligned}
&x^3+ x^2 y + xy^2 + y^3 = 81\\
Sol^n\\
&Differentiating \quad w.r.t. \quad x\\
&\frac{d}{d x}\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)=\frac{d}{d x}81\\
&\frac{d\left(x^{3}\right)}{d x}+\frac{d\left(x^{2} y\right)}{d x}+\frac{d\left(x y^{2}\right)}{d x}+\frac{d\left(y^{3}\right)}{d x}=0 \\
&3 x^{2}+\frac{d\left(x^{2} y\right)}{d x}+\frac{d\left(x y^{2}\right)}{d x}+3 y^{2} \frac{d y}{d x}=0\\
&3 x^{2}+\left(y\frac{d}{dx}x^2 +x^{2} \cdot \frac{d y}{d x }\right)+\left(x\frac{d}{d x} y^{2}+y^2\frac{d}{dx} x\right)+3 y^{2} \frac{d y}{y}=0\\
&3 x^{2}+\left(2 x \cdot y+x^{2} \frac{d y}{d x}\right)+\left(x 2 y \frac{d y}{d x}+ y^2 \right)+3 y^{2} \frac{d y}{d x}=0\\
&3 x^{2}+2 x y+x^{2} \frac{d y}{d x}+y^2+2 xy\frac{d y}{d x}+3 y^{2} \frac{d y}{d x}=0\\
&x^{2} \frac{d y}{d x}+2 xy\frac{d y}{d x}+3 y^{2} \frac{d y}{d x}=-3 x^{2}-2 x y-y^2\\
&\frac{d y}{d x} [x^{2} +2 xy+3 y^{2}]=-3 x^{2}-2 x y-y^2\\
&\frac{d y}{d x} =-\frac{3 x^{2}+2 x y+y^2}{x^{2} +2 xy+3 y^{2}}\end{aligned}$
v)
$\begin{aligned}
&x^{2} y^{2}-\tan ^{-1} \sqrt{x^{2}+y^{2}}=\cot ^{-1} \sqrt{x^{2}+y^{2}}\\
&x^{2} y^{2}=\tan ^{-1} \sqrt{x^{2}+y^{2}}+\cot ^{-1} \sqrt{x^{2}+y^{2}}\\
&Using \quad \tan ^{-1} \theta+\cot ^{-1} \theta=\frac{\pi}{2}\\
&\therefore \quad x^{2} y^{2}=\frac{\pi}{2}\\
&Take \quad squartroot \quad both \quad side\\
&\sqrt{x^{2} y^{2}}=\sqrt{\frac{\pi}{2}}\\
&x y=\sqrt{\frac{\pi}{2}}\\
&Diff \quad w.r.t\quad x \\
&x \cdot \frac{d y}{d x}+y \frac{d}{d x}(x)=\frac{d}{d x} \sqrt{\frac{\pi}{2}}\\
&x \cdot \frac{d y}{d x}+y=0\\
&\frac{d y}{d x}=\frac{-y}{x}\end{aligned}$
vi)
$\begin{aligned}
&xe^y + ye^x = 1\\
sol^n\\
&\left(x \frac{d}{d x}e^{y} +e^{y} \frac{d}{d x} x\right)+\left(y \frac{d}{d x}e^{x}+e^{x} \cdot \frac{d y}{d x}\right)=0\\
&\left(x .e^{y} \frac{d y}{d x}+e^{y} .1\right)+\left(y e^{x}+e^{x} \cdot \frac{d y}{d x}\right)=0\\
&\frac{d y}{d x}\left(x e^{y}+e^{x}\right)=-e^{y}-y e^{x}\\
&\Rightarrow \frac{d y}{d x}=-\frac{e^{y}+y e^{x}}{x e^{y}+e^{x}}
\end{aligned}$
vii)
$\begin{aligned}
&e^{x+y}=\cos (x-y) \\
&Diff \quad w.r.t \quad x\\
&\frac{d}{d x} e^{x+y}=\frac{d}{d x} \cos (x-4) \\
&e^{x+4} \frac{d}{d x}(x+y)=-\sin (x-4) \cdot \frac{d}{d x}(x-y) \\
&e^{x+4} \cdot\left(1+\frac{d y}{d x}\right)=-\sin (x-4) \cdot\left(1-\frac{d y}{d x}\right)\\
&e^{x+y}+e^{x+y} \cdot \frac{d y}{d x}=-\sin (x-y)+\sin (x-4) \frac{d y}{a x}\\
&e^{x+y} \frac{d y}{d x}-\sin (x-4) \frac{d y}{d x}=-\sin (x-4)-e^{x+y}\\
&\frac{d y}{d x}\left[e^{x+4}-\sin (x-4)\right]=-\left[\sin (x-4)+e^{x+y}\right]\\
&\frac{d y}{d x}=-\left[\frac{e^{x+4}+\sin (x-y)}{e^{x+y}-\sin (x-y)}\right]
\end{aligned}$
viii)
$\begin{aligned}
&\cos (x y)=x+y\\
&Diff w.r.t \quad x\\
&\frac{d}{d x} \cos (x y)=\frac{d}{d x}(x+y)\\
&-\sin (x y)\left[x \cdot \frac{d y}{d x}+y \cdot \frac{d}{d x} x\right]=1+\frac{d y}{d x}\\
&-\sin (xy)\left[x \cdot \frac{d y}{d x}+y\right]=1+\frac{d y}{d x}\\
&-\sin (x y) x \cdot \frac{d y}{d x}-y\sin (x y)=1+\frac{d y}{d x}\\
&\frac{d y}{d x}[-x \sin (x y)-1]=1+y \sin (x y)\\
&\frac{d y}{d x}[-(x \sin (x y)+1)]=1+y \sin (x y)\\
&\frac{d y}{d x}=-\frac{1+y \sin (x y)}{1+x \sin (x y)}\end{aligned}$
ix)
$\begin{aligned}
&e^{e^{x-y}}=\frac{x}{y}\\
&Taking \quad log \quad on \quad both \quad side\\\\
&\log e^{e^{x-y}}=\log \left(\frac{x}{y}\right)\\
&e^{x-y} \log e=\log x-\log y\\
&e^{x-y}=\log x-\log y\\\\
&Diff \quad w.r.t \quad x\\
&\frac{d}{d x} e^{x-y}=\frac{d}{d x} \log x-\frac{d}{d x} \log y\\
&e^{x-y} \cdot \frac{d}{d x}(x-y)=\frac{1}{x}-\frac{1}{y} \cdot \frac{d y}{d x}\\
&e^{x-y} \cdot\left(1-\frac{d y}{d x}\right)=\frac{1}{x}-\frac{1}{y} \frac{d y}{d x}\\
&e^{x-y}-e^{x-y} \frac{d y}{d x}=\frac{1}{x}-\frac{1}{y} \frac{d y}{d x}\\
&\frac{1}{y} \frac{d y}{d x}-e^{x-y} \frac{d y}{d x}=\frac{1}{x}-e^{x-y}\\
&\frac{d y}{d x}\left[\frac{1}{y}-e^{x-y}\right]=\frac{1-x e^{x-y}}{x}\\
&\frac{d y}{d x}\left[\frac{1-y e^{x-y}}{y}\right]=\frac{1-x e^{x-y}}{x}\\
&\frac{d y}{d x}=\frac{y\left(1-x e^{x-y}\right)}{x\left(1-y e^{x-y}\right)}\end{aligned}$
x)
$\begin{aligned}
&x+\sin (x+y)=y-\cos (x-y)\\
&Diff \quad w.r.t \quad x\\
&\frac{d}{d x} x+\frac{d}{d x} \sin (x+y)=\frac{d y}{d x}-\frac{d}{d x} \cos (x-y)\\
&1+\cos (x+y) \cdot \frac{d}{d x}(x+y)=\frac{d y}{d x}+\sin (x-y) \frac{d}{d x}(x-y)\\
&1+\cos (x+y)\left(1+\frac{d y}{d x}\right)=\frac{d y}{d x}+\sin (x-4)\left(1-\frac{d y}{d x}\right)\\
&1+\cos (x+4)+\cos (x+4) \cdot \frac{d y}{d x}=\frac{d y}{d x}+\sin (x-4)-\sin (x+1)\frac {d y}{d x}\\
&\frac{d y}{d x}[\cos (x+y)+\sin (x-4)-1]=\sin (x-4)-\cos (x+y)-1\\
&\frac{d y}{d x}=\frac{\sin (x-4)-\cos (x+4)-1}{\cos (x+4)+\sin (x-4)-1}\end{aligned}$
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