H.S.C.
Differentiation EX 1.1 Q.3 Maharashtra state Board Mathematics and statistic for arts and science
Differentiation EX 1.1
Q.3
i)
ii)
iii)
iV)Coming soon !!
V)
Vi)
Q.3
i)
$\begin{array}{rl}
y&=\left(x^{2}+4 x+1\right)^{3}+\left(x^{3}-5 x-2\right)^{4}\\
&Diff w.r.t x\\
\frac{d y}{d x}&=\frac{d}{d x}\left[\left(x^{2}+4 x+1\right)^{3}+\left(x^{3}-5 x-2\right)^{4}\right]\\
&=\frac{d}{d x}\left(x^{2}+4 x+1\right)^{3}+\frac{d}{d x}\left(x^{3}-5 x-2\right)^{4}\\
&=3 \cdot\left(x^{2}+4 x+1\right)^{2}\frac{d}{d x}\left(x^{2}+4 x+1\right)^{4}+4\left(x^{3}-5 x-2\right)^{3} \cdot \frac{d}{d x}\left(x^{3}-5 x-2\right)\\
&=3\left(x^{2}+4 x+1\right)^{2} \cdot(2 x+4)+4\left(x^{3}-5 x-2\right)^{3} \cdot\left(3 x^{2}-5\right)\\
&=6\left(x^{2}+4 x+1\right)^{2} \cdot(x+2)+4\left(3 x^{2}-5\right)\left(x^{3}-5 x-2\right)^{3}\\
&=6\left(x+2)\cdot(x^{2}+4 x+1\right)^{2} +4\left(3 x^{2}-5\right)\left(x^{3}-5 x-2\right)^{3}\end{array}$
ii)
$\begin{array}{rl}
y&=(1+4 x)^{5}\left(3+x-x^{2}\right)^{8}\\
Using\\
y&=u \cdot v\\
\frac{d y}{d x}&=u \cdot \frac{d v}{d x}+v \cdot \frac{d u}{d x}\\
&Diff.. w. x . t \quad x \\
\frac{d y}{d x}&=(1+4 x)^{5} \cdot \frac{d}{d x}\left(3+x-x^{2}\right)^{8}+\left(3+x-x^{2}\right)^{8} \frac{d}{d x}(1+4 x)^{5}\\
&=(1+4 x)^{5} \cdot 8\left(3+x-x^{2}\right)^{7} \frac{d}{d x}\left(3+x-x^{2}\right)+\left(3+x-x^{2}\right)^{8} \cdot 5(1+4 x) \cdot \frac{d}{d x}(1+4 x)\\
&=(1+4 x)^{5} \cdot 8\left(3+x-x^{2}\right)^{7} \cdot(0+1-2 x)+\left(3+x-x^{2}\right)^{8} \cdot 5(1+4 x)(4)\\
&\left.=8(1-2 x)(1+4 x)^{5}(3+x-x^{2}\right)^{7}+20(1+4 x)\left(3+x-x^{2}\right)^{8}\end{array}$
iii)
$\begin{array}{rl}
y&=\frac{x}{\sqrt{7-3 x}}\\
Using \\
y&=\frac{u}{v} \\
\frac{d y}{d x}&=\frac{v \cdot d u}{d x}-u \cdot \frac{d v}{d x}\\
&diff \quad y \quad w.r.t \quad x \\
\frac{d y}{d x}&=\frac{d}{d x}\left(\frac{x}{\sqrt{7-3 x}}\right)\\
&=\frac{\sqrt{7-3 x} \cdot \frac{d}{d x}(x)-x \cdot \frac{d}{d x} \sqrt{7-3 x}}{[\sqrt{7-3 x}]^{2}}\\
&=\frac{\sqrt{7-3 x} \cdot-\frac{1}{2\sqrt{7-3 x}} \cdot \frac{d}{d x}(7-3 x)}{(7-3 x)}\\
&=\frac{\sqrt{7-3 x}-\frac{x}{2\sqrt{7-3 x}} \cdot(-3)}{(7-3 x)}\\
&=\frac{\sqrt{7-3 x}+\frac{3 x}{2\sqrt{7-3x}}}{7-3 x^{2}}\\
&=\frac{2{(7-3 x)+3x}}{2(7-3 x^{2})^\frac{3}{2}}\\
&=\frac{{14-6x+3x}}{2(7-3 x^{2})^\frac{3}{2}}\\
&=\frac{{14-3x}}{2(7-3 x^{2})^\frac{3}{2}}\end{array}$
iV)Coming soon !!
V)
$\begin{array}{rl}
y&=\left(1+\sin ^{2} x\right)^{2}\left(1+\cos ^{2} x\right)^{3}\\
y&=u\\
\frac{d y}{d x}&=u \cdot \frac{d v}{d x}+v \cdot \frac{d u}{d x}\\
&Diff.w.r.t \quad x\\
\frac{d y}{d x}&=\left(1+\sin ^{2} x\right)^{2} \frac{d}{d x}\left(1+\cos ^{2} x\right)^{3}+\left(1+\cos ^{2} x\right)^{3} \frac{d}{d x}\left(1+\sin ^{2} x\right)^{2}\\
&=\left(1+\sin ^{2} x\right)^{2} \cdot 3\left(1+\cos ^{2} x\right)^{2} \frac{d}{d x}\left(1+\cos ^{2} x\right)+\left(1+\cos ^{2} x\right)^{3} \cdot 2\left(1+\sin ^{2} x\right) \cdot \frac{d}{d x}\left(1+\sin ^{2} x\right)\\
&=\left(1+\sin ^{2} x\right)^{2} \cdot 3\left(1+\cos ^{2} x\right)^{2} \cdot 2 \cdot \cos x(-\sin x)+\left(1+\cos ^{2} x\right)^{3} \cdot 2\left(1+\sin ^{2} x\right) \cdot 2 \cdot \sin x \cdot \cos x\\
&=\left(1+\sin ^{2} x\right)^{2} \cdot 3\left(1+\cos ^{2} x\right)^{2}(-\sin 2 x)+\left(1+\cos ^{2} x\right)^{3}2\left(1+\sin ^{2} x\right) \cdot \sin 2 x\\
&=\left(1+\sin ^{2} x\right)\cdot\left(1+\cos ^{2} x\right)^{2}(\sin 2 x)\left[-3\left(1+\sin ^{2} x\right)+2\left(1+\cos ^{2} x\right)\right]\\
&=\left(1+\sin ^{2} x\right)\left(1+\cos ^{2} x\right)^{2} \cdot \sin 2 x\left[-3-3 \sin ^{2} x+2+2 \cos ^{2} x\right]\\
&=\left(1+\sin ^{2} x\right)\left(1+\cos ^{2} x\right)^{2} \cdot \sin 2 x\left[-1-3 \sin ^{2} x+2\left(1-\sin ^{2} x\right)\right]\\
&=\left(1+\sin ^{2} x\right)\left(1+\cos ^{2} x\right)^{2} \cdot \sin 2 x\left(-1-3 \sin ^{2} x+2-2\sin ^{2} x\right)\\
&=\left(1+\sin ^{2} x\right)\left(1+\cos ^{2} x\right)^2 \cdot \sin 2 x\left(1-5 \sin ^{2} x\right)\\
&=\sin 2 x\left(1+\sin ^{2} x\right)\left(1+\cos ^{2} x\right)^2 \left(1-5 \sin ^{2} x\right)\end{array}$
Vi)
$\begin{array}{rl}
&\text { vi) } y=\sqrt{\cos x}+\sqrt{\cos \sqrt{x}}\\
&\text { Diff }w.r.t \quad x\\
&\frac{d y}{d x}=\frac{d}{d x} \sqrt{\cos x}+\frac{d}{d x} \sqrt{\cos \sqrt{x}}\\
&=\frac{1}{2 \sqrt{\cos x}} \cdot \frac{d}{d x} \cos x+\frac{1}{2 \sqrt{\cos \sqrt{x}}} \cdot \frac{d}{d x} \cos \sqrt{x}\\
&=\frac{1}{2 \sqrt{\cos x}}(-\sin x)+\frac{1}{2 \sqrt{\cos \sqrt{x}}}(-\sin \sqrt{x})\frac{d}{d x} \sqrt{x}\\
&=\frac{1}{2 \sqrt{\cos x}}(-\sin x)-\frac{\sin \sqrt{x}}{2 \sqrt{\cos \sqrt{x}}} \times \frac{1}{2 \sqrt{x}}\\
&=\frac{-\sin x}{2 \sqrt{\cos x}}-\frac{\sin \sqrt{x}}{4 \sqrt{x} \sqrt{\cos \sqrt{x}}}
\end{array}$
Vii)
Viii)Coming soon !!
iX)
Xv)
Vii)
$\begin{array}{rl}
y&=\log (\sec 3 x+\tan 3 x)\\
&Diff \quad w.r.t. \quad x\\
\frac{d y}{d x}&=\frac{d}{d x} \log (\sec 3 x+\tan 3 x)\\
&=\frac{1}{\sec 3 x+\tan 3 x} \cdot \frac{d}{d x}(\sec 3 x+\tan 3 x)\\\\
&=\frac{1}{\sec 3 x+\tan 3 x} \cdot\left[\sec 3 x \cdot \tan 3 x \cdot \frac{d}{d x} 3 x+\sec ^{2} 3 x\right.\left.\frac{d}{d x} \cdot 3 x\right]\\\\
&=\frac{1}{\sec 3 x+\tan 3 x}\left[3 \cdot \sec 3 x \cdot \tan x+3 \sec ^{2} 3 x\right]\\\\
&=\frac{3 \sec 3 x[\tan x+\sec 3 x]}{(\sec 3 x+\tan 3 x]}\\\\
&=3 \sec 3 x\end{array}$
Viii)Coming soon !!
iX)
------
$\begin{array}{rl}
y&=\cot \left(\frac{\log x}{2}\right)-\log \left(\frac{\cot x}{2}\right)\\
&Diff \quad w.r.t \quad x \\
\frac{d y}{d x}&=\frac{d}{d x} \cot \left(\frac{\log x}{2}\right)-\frac{d}{d x} \log \left(\frac{\cot x}{2}\right)\\
&=-\operatorname{cosec}^{2}\left(\frac{\log x}{2}\right) \cdot \frac{d}{d x}\left(\frac{\log x}{2}\right)-\frac{1}{\frac{\cot x}{2}} \times \frac{d}{d x}\left(\frac{\cot x}{2}\right)\\\\
&=-cosec^{2}\left(\frac{\log x}{2}\right) \frac{1}{2} \times \frac{1}{x}-\frac{1}{\left(\frac{\cot x}{2}\right)} {(-cosec^{2}x)} \frac{1}{2}\\\\
&=\frac{-\operatorname{cosec}^{2}\left(\frac{\log x}{2}\right)}{2 x}-\frac{2\tan x}{2} .\operatorname{cosec}^{2} x\\
&=\frac{-\operatorname{cosec}^{2}\left(\frac{\log x}{2}\right)}{2}+(\tan x)\left(\operatorname{cosec}^{2} x\right)\end{array}$
X)
$\begin{array}{rl}
y&=\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}\\
Using \\
y&=\frac{u}{v} \\ \frac{d y}{d x} & =\frac{v \cdot d u}{d x}-u \cdot \frac{d v}{d x}\\
&Diff \quad w.r.t\quad x\\
&\frac{d y}{d x}=\frac{\left(e^{2 x}+e^{-2 x}\right) \cdot \frac{d}{d x}\left(e^{2 x}-e^{-2 x}\right)-[e^{2 x}-e^{-2 x} \cdot \frac{d}{d x}\left(e^{2 x}+e^{-2 x}\right)]}{\left[e^{2 x}+e^{-2 x}\right]^{2}}\\
&=\frac{[\left(e^{2 x}+e^{-2 x}\right)\left(e^{2 x} \cdot 2-e^{-2 x} \cdot(-2)\right]-\left[\left(e^{2 x}-e^{-2 x}\right) \cdot\left(e^{2 x} \cdot 2+e^{-2 x} \cdot(-2)]\right.\right.}{\left[e^{2 x}+e^{-2 x}\right]^{2}}\\
&=\frac{\left(e^{2 x}+e^{-2 x}\right)\left(2 e^{2 x}+2 e^{-2 x}\right)-\left[\left(e^{2 x}-e^{-2 x}\right)\left(2 e^{2 x}-2 e^{-2 x}\right)\right]}{\left[e^{2 x}+e^{-2 x}\right)^{2}}\\
&=\frac{\left[2 e^{4 x}+2 e^{2 x} \cdot e^{-2 x}+2 e^{2 x} \cdot e^{-2 x}+2 e^{-4 x}\right]-\left[2 e^{4 x}-2 e^{-2 x} e^{2 x}-2 e^{2 x}e^{-2 x}+e^{-4 x} \right]1}{(e^{2 x}+e^{-2 x})^{2})}\\
&=\frac{2 e^{4 x}+2 e^{0}+2 e^{0}+2 e^{-4 x} - e^{4 x}+2 e^{0}+2 e^{0}-2 e^{-4x}}{(e^{2 x}+e^{-2 x})^{2}}\\
&=\frac{8}{\left[e^{2 x}+e^{-2 x}\right]^{2}}\\
&=\frac{8}{\left[e^{2 x}+\frac{1}{e^{2 x}}\right]^{2}}\\
&=\frac{8\left(e^{2 x}\right)^{2}}{\left(e^{4 x}+1\right)^{2}}\\
&=\frac{8 e^{4 x}}{\left(e^{4 x}+1\right)^{2}}\end{array}$
Xi)
$\begin{array}{rl}
y&=\frac{e^{\sqrt{x}}+1}{\sqrt{x}-1}\\
&Using \\
y&=\frac{u}{v} \\
\frac{d y}{d x}&=\frac{v \cdot \frac{d u}{d x}-u \cdot \frac{d v}{d x}}{v^{2}}\\
&Diff\quad wrt \quad x \frac{d y}{d x}\\ &=\frac{\left(e^{\sqrt{x}}-1\right) \cdot \frac{d}{d x}\left(e^{\sqrt{x}}+1\right)-\left(e^{\sqrt{x}}+1\right) \cdot \frac{d}{d x}\left(e^{\sqrt{x}}-1\right)}{\left(e^{\sqrt{x}}-1\right)^{2}} \\ &=\frac{\left(e^{\sqrt{x}}-1\right) \cdot e^{\sqrt{x}} \times \frac{1}{2 \sqrt{x}}-\left(e^{\sqrt{x}}+1\right) \cdot e^{\sqrt{x}} \times \frac{1}{2 \sqrt{x}}}{\left(e^{\sqrt{x}}-1\right)^{2}} \\ &=\frac{\left(e^{\sqrt{x}}-1\right) e^{\sqrt{x}}-e^{\sqrt{x}}\left(e^{\sqrt{x}}+1\right)}{2 \sqrt{x}\left(e^{\sqrt{x}}-1\right)^{2}} \\ &=\frac{e^{2 \sqrt{x}}-e^{\sqrt{x}}-e^{2 x}-e^{\sqrt{x}}}{2 \sqrt{x} \cdot\left(e^{-x}-1\right)^{2}} \\ &=\frac{-2 e^{\sqrt{x}}}{2 \sqrt{x}\left(e^{\sqrt{x}}-1\right)^{2}} \\ &=\frac{-e^{\sqrt{x}}}{\sqrt{x}\left(e^{\sqrt{x}}-1\right)^{2}}
\end{array}$
Xii)
$\begin{array}{rl}
y&=\log \left[\tan ^{3} x \sin ^{4} x \cdot\left(x^{2}+7\right)^{7}\right]\\
&Using \\
&[\log (A \cdot B \cdot C)=\log A+\log B+\log C \\
&\log a^{b}=b \log a]\\
y&=\log \tan ^{3} x+\log \sin ^{4} x+\log \left(x^{2}+7\right)^{7}\\
& Diff \quad w.r.t\quad x\\
\frac{d y}{d x}&=\frac{d}{d x} \log \left(\tan ^{3} x\right)+\frac{d}{d x} \log \sin ^{4} x \cdot \log \left(x^{2}+7\right)^{7}\\
\frac{d y}{d x}&=\frac{d}{d x} 3 \log (\tan x)+\frac{d}{d x} 4 \log \sin x+\frac{d}{d x} 7 \cdot \log \left(x^{2}+7\right)\\
&=\frac{3}{\tan x} \sec ^{2} x+\frac{4}{\sin x} \cdot \cos x+\frac{7}{x^{2}+7} 2x\\
&=\frac{3 \cos x}{\sin x} \frac{1}{\cos ^{2} x}+4 \cdot \cot x+\frac{14 x}{x^{2}+7}\\
&=\frac{3 }{\sin x} \frac{1}{\cos x}+4 \cdot \cot x+\frac{14 x}{x^{2}+7}\\
&=\frac{2 \times3 }{2 \times \sin x} \frac{1}{\cos x}+4 \cdot \cot x+\frac{14 x}{x^{2}+7}\\
&=\frac{2 \times3 }{\sin {2} x}+4 \cot x+\frac{14 x}{x^{2}+7} \quad [\because 2sinx.coxx=sin2x]\\
&=6 \cdot \operatorname{cosec} 2 x+4 \cot x+\frac{14 x}{x^{2}+7}\\
\end{array}$
Xiii)
$\begin{array}{rl}
y&=\log (\sqrt{\frac{1-\cos 3 x}{1+\cos 3 x}})\\
&using \\
&\log \left(\frac{a}{b}\right)=\log a-\log b\\
y&=\log (\sqrt{1-\cos 3 x})-\log (\sqrt{1+\cos 3} x)\\
&=\log (1-\cos 3 x)^{1 / 2}-\log (1+\cos 3 x)^{1 / 2}\\
y&=\frac{1}{2} \log (1-\cos 3 x)-\frac{1}{2} \log (1+\cos 3 x)\\
&Diff\quad w.r.t. \quad x\\
\frac{d y}{d x}&=\frac{1}{2} \frac{1}{(1-\cos 3 x)} \frac{d}{d x}(1-\cos 3 x)-\frac{1}{2} \cdot \frac{1}{1+\cos 3 x} \cdot \frac{d}{d x}(1+\cos 3 x)\\
&=\frac{1}{2(1-\cos 3 x)}(\sin 3 x \cdot(3))-\frac{-\sin 3 x \cdot(3)}{2(1+\cos 3 x)}\\
&=\frac{3 \sin 3 x}{2(1-\cos 3 x)}+\frac{3 \sin 3 x}{2(1+\cos 3 x)}\\
&=\frac{3}{2}\left[\frac{\sin 3 x}{1-\cos 3 x}+\frac{\sin 3 x}{1+\cos 3 x}\right]\\
&\frac{3}{2}\left[\frac{\sin 3 x(1+\cos 3 x)+\sin 3 x(1-\cos 3 x)}{\left(1-\cos ^{2} 3 x\right)}\right]\\
&=\frac{3}{2}\left[\frac{\sin 3 x+\sin 3 x \cdot \cos 3 x+\sin 3 x-\sin 3 x \cdot \cos 3 x}{\left(1-\cos ^{2} 3 x\right)}\right]\\
&=\frac{3}{2}\left[\frac{2 \cdot \sin 3 x}{\sin ^{2} 3 x}\right] \quad [\because 1-\cos ^{2} x=\sin ^{2} x]\\
&=\frac{3}{2} \times 2 \times \frac{\sin 3 x}{\sin ^{2} 2 x}\\
&={3}\frac{1}{\sin 2 x}\\
&=3\operatorname{cosec} 3 x\end{array}$
Xiv)
$\begin{array}{rl}
y&=\log [\sqrt{\frac{1+\cos \left(\frac{5 x}{2}\right)}{1-\cos \left(\frac{5 x}{2}\right)}}\\
Using\\
\log \left(\frac{a}{b}\right)&=\log a-\log b\\
\log a^{b}&=b \log a\\
y&=\log (\sqrt{1+\cos \frac{5 x}{2}})-\log (\sqrt{1-\cos \left(\frac{5x}{2}\right)})\\
y&=\log \left(1+\cos \left(\frac{x}{2}\right)^{1 / 2}-\log \left(1-\cos \left(\frac{5 x}{2}\right)\right)^{1 / 2}\right.\\
y&=\frac{1}{2} \log \left[1+\cos \left(\frac{5 x}{2}\right)\right]-\frac{1}{2} \log \left[\left(1-\cos \left(\frac{5 x}{2}\right)\right]\right.\\
&Diff \quad w.r.t. \quad x \\
\frac{d y}{d x}&=\frac{1}{2} \frac{1}{1+\cos \left(\frac{5 x}{2}\right)} \frac{d}{d x}\left(1+\cos \frac{5 x}{2}\right)-\frac{1}{2} \times \frac{1}{1-\cos \left(\frac{5 x}{2}\right)} \frac{d}{d x}\left(1-\cos \frac{5 x}{2}\right)\\
&=\frac{1}{2\left(1+\cos \left(\frac{5 x}{2}\right)\right.} (-\sin \left(\frac{5 x}{2}\right) \cdot \frac{5}{2}-\frac{1}{2\left(1-\cos \left(\frac{5 x}{2}\right))\right.} \left(\sin \left(\frac{5 x}{2}\right) \cdot \frac{5}{2}\right.\\
&=\frac{-5\sin \left(\frac{5 x}{2}\right)}{4\left(1+\cos \left(\frac{5 x}{2}\right)\right)}-\frac{5 \sin \left(\frac{5 x}{2}\right)}{4\left(1-\cos \left(\frac{5 x}{2}\right)\right]}\\
&=\frac{-5}{4} \sin \left(\frac{5 x}{2}\right)\left[\frac{1}{1+\cos \left(\frac{5 x}{2}\right)}+\frac{1}{1-\cos \left(\frac{5 x}{2}\right)}\right]\\
&\frac{\frac{-5}{4} \sin (5 x/2)\left[1-\cos \left(\frac{5 x}{2}\right)+1+\cos \frac{5 x}{2}\right]}{\left[1-\cos ^{2} (\frac{5 x}{2})\right]}\\
&={\frac{-5}{4}}\sin \left(\frac{5 x}{2}\right) \times \frac{2}{\sin ^{2} (\frac{5 x}{2})} \quad [\because 1-\cos ^{2} x=\sin ^{2} x]\\
&={\frac{-5}{4}}\frac{1}{\sin(\frac{5 x}{2})} \\
&-\frac{5}{2} \times cosec x
\end{array}$
Xv)
$\begin{aligned} y &=\log (\sqrt{\frac{1-\sin x}{1+\sin x}}\\
y &=\log (\sqrt{\frac{1-\sin x}{1+\sin x} \times \frac{1-\sin x}{1-\sin x}}\\
&=\log (\sqrt{\frac{(1-\sin x)^{2}}{1-\sin ^{2} x}}) \quad [\because\sin ^{2} x=\cos ^{2} x ] \\
&=\log \sqrt{\frac{(1-\sin x)^{2}}{\cos ^{2} x}}\\
&=\log \sqrt{\left(\frac{1-\sin x}{\cos x}\right)^{2}} \\
y &=\log \left(\frac{1-\sin x}{\cos x}\right) \\
y &=\log (1-\sin x)-\log \cos x \\
&Diff. w.r.t. \quad x\\
\frac{d y}{d x}&=\frac{1}{(1-\sin x)} \cdot \frac{d}{d x}(1-\sin x)-\frac{1}{\cos x} \cdot \frac{d}{d x} \cos x\\
&=\frac{1}{(1-\sin x)} (-\cos x)-\frac{1}{\cos x}(-\sin x)\\
&=\frac{-\cos x}{1-\sin x}+\frac{\sin x}{\cos x}\\
&=\frac{\left(-\cos ^{2} x+\sin x-\sin ^{2} x\right)}{\cos x(1-\sin x)}\\
&=\frac{\left[\sin x-\left(\sin ^{2} x+\cos ^{2} x\right)\right]}{\cos x(1-\sin x)}\\
&=\frac{(\sin x-1)}{\cos x(1-\sin x)}\quad \\
&=\frac{-(1-\sin x)}{\cos x(1-\sin x)}\\
&=\frac{-1}{\cos x}\\
&=-\sec x\end{aligned}$
Xvi)
$\begin{aligned}
&y=\log \left[4^{2 x}\left(\frac{x^{2}+5}{\sqrt{2 x^{3}-4}}\right)^{\frac{3}{2}}\right]\\
&using\\
[&\log(A.B)=\log A+\log B\\
&\log(\frac{A}{B})=\log A-\log B \\
&\log A^B=B\log A]\\\\
&y=\log 4^{2 x}+\log \left(\frac{x^{2}+5}{\sqrt{2 x^{3}-4}}\right)^{3 / 2}\\
&y=2 x \cdot \log 4+\frac{3}{2} \log \left(\frac{x^{2}+5}{\sqrt{2 x^{3}-4}}\right)\\
&y=2 x \log 4+\frac{3}{2}\left[\log \left(x^{2}+5\right)-\log (\sqrt{2 x^{3}-4})\right]\\
&\begin{array}{l}
y=2 x \log{4}+\frac{3}{2} \log \left(x^{2}+5\right)-\frac{3}{2} \times \frac{1}{2} \log \left(2 x^{3}-4\right) \\
\text { Diff } w.x.t \quad x
\end{array}\\
&\frac{d y}{d x}=\log4 \frac{d}{d x} 2 x+\frac{3}{2} \frac {d}{d x}\log \left(x^{2}+5\right)-\frac{3}{4} \frac{d}{d x} \log \left(2 x^{3}-4\right)\\
&=\log 4 . 2+\frac{3}{2} \cdot \frac{1}{x^{2}+5} 2x-\frac{3}{4} \frac{1}{2 x^{3}-4} 2\times 3 x^{2}\\
&=2 \log 4+\frac{3 x}{x^{2}+5}-\frac{9 x^{2}}{2\left(2 x^{3}-4\right)}
\end{aligned}$
Xvii)
$\begin{aligned}
y&=\log \left[\frac{e^{x^{2}}(5-4 x)^{3 / 2}}{\sqrt[3]{7-6 x}}\right]\\
&Using\\
&\log(A.B)=\log A+\log B\\
y&=\log e^{x^{2}}+\log \left(\frac{(5-4 x)^{3 / 2}}{\sqrt[3]{7-6 x}}\right) \\
&=\log e^{x^{2}}+\log (5-4 x)^{3 / 2}-\log (\sqrt[3]{7-6 x})\\
&=x^{2} \log e+\frac{3}{2} \log (5-4 x)-\log (7-6 x)^{1 / 3}\\
&=x^{2}+\frac{3}{2} \log (5-4 x)-\frac{1}{3} \log (7-6 x)\\
&NOW,\\
&Diff w.r.t \quad x\\
\frac{d y}{d x} &=\frac{d}{d x} x^{2}+\frac{3}{2}\frac{d}{d x} \log(5-4 x)- \frac{1}{3}\frac{d}{d x} \log (7-6 x)\\
\\ &=2 x+\frac{3}{2} \frac{1}{5-4 x} (-4)-\frac{1}{3} \frac{1}{(7-6 x)} x(-6) \\
&=2 x-\frac{6}{(5-4 x)}+\frac{2}{(7-6 x)}\end{aligned}$
Xviii)
$\begin{aligned}
y&=\log \left[\frac{a^{\log x}}{\left(x^{2}-3\right)^{3} \log x}\right]\\
y&=\log a^{\cos x}-\log \left(\left(x^{2}-3\right)^{3} \cdot \log x\right]\\
y&=\cos x \cdot \log a-\left[\log \left(x^{2}-3\right)^{3}+\log (\log x)\right]\\
y&=\cos x \cdot \log a-3 \log \left(x^{2}-3\right)-\log (\log x)\\
&Diff \quad w.r.t \quad x\\
\frac{d y}{d x}&=\log a\frac{d}{d x}(\cos x) -3 \cdot \frac{d}{d x} \log \left(x^{2}-3\right)-\frac{d}{d x} \log (\log x)\\
&=(-\sin x) \log a-3 \frac{1}{(x^{2}-3)}.2x-\frac{1}{\log x} \times \frac{1}{x}\\
&=-\sin x \cdot \log a-\frac{6 x}{\left(x^{2}-3\right)}-\frac{1}{x \cdot \log x}\\
\end{aligned}$
Xix)
$\begin{aligned}
y&=(25)^{\log _{5}(\sec x)}-16^{\log _{4}(\tan x)}\\
y&=\left(5^{2}\right)^{\log _{5}(\sec x)}-\left(4^{2}\right)^{\log _{4}(\tan x)}\\
y&=5^{2 \log _{5}(\sec x)}-4^{2 \log _{4} \tan x}\\
y&=5^{\log _{5}\left(\sec ^{2} x\right)}-4^{\log _{4} \tan ^{2} x}\\
Using \\
[e^{\log _{e}}&=1 \\
5^{\log _{5}}&=1]\\
y&=\sec ^{2} x-\tan ^{2} x\\
y&=1+\tan ^{2} x-\tan ^{2} x\\
y&=1\\
&Diff \quad w.r.t. \quad x\\
\frac{d y}{d x}&=0 \end{aligned}$
XX)
*Note Answer is incorrect in Book
$\begin{aligned}
y&=\frac{\left(x^{2}+2\right)^{4}}{\sqrt{x^{2}-5}}\\
using \\
y&=\frac{u}{v}\\
\frac{d y}{d x}&=\frac{v \cdot \frac{d u}{d x}-u \cdot \frac{d v}{d x}}{v^{2}}\\\\
&Diff w.r.t \\
\frac{d y}{d x}&=\frac{\sqrt{x^{2}-5} \cdot \frac{d}{d x}\left(x^{2}+2\right)^{4}-\left(x^{2}+2\right)^{4} \cdot \frac{d}{d x} \cdot \sqrt{x^{2}-5}}{[\sqrt{x^{2}-5})^{2}]}\\
&=\frac{\sqrt{x^{2}-5} \cdot 4\left(x^{2}+2\right)^{3} \cdot 2 x-\left(x^{2}+2\right)^{4} \frac{1}{2\sqrt {x^{2}-5}}.2x}{\left(x^{2}-5 \right)}\\
&=\frac{8 x\left(x^{2}+2\right)^3(\sqrt{x^{2}-5})-\frac{x\left(x^{2}+2\right)^{4}}{\sqrt{x^{2}-5}}}{\left(x^{2}-5\right)}\\
&=\frac{8 x\left(x^{2}+2\right)^{3}\left(x^{2}-5\right)-x\left(x^{2}+2\right)^{4}}{\left(x^{2}-5\right) \cdot\left(x^{2}-5\right)^{1 / 2}}\\
&=\frac{x\left(x^{2}+2\right)^{3}\left[8\left(x^{2}-5\right)-\left(x^{2}+2\right)\right]}{\left(x^{2}-5\right)^{3 / 2}}\\\\
&=\frac{x\left(x^{2}+2\right)^{3}(8x^2-40-x^2-2)}{\left(x^{2}-5\right)^{3 / 2}}\\
&=\frac{x\left(x^{2}+2\right)^{3}(7x^2-42)}{\left(x^{2}-5\right)^{3 / 2}}\end{aligned}$
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