Differentiate the following w.r.t x
1)$y=log_3(log_5 x)$
Solution-
$\therefore$ y=$\frac{log(log x)}{log 3}$-$log_3(log 5)$
$\frac{dy}{dx}$=$\frac{1}{log 3}$$\frac{d}{dx}[log(logx)]$-$\frac{d}{dx}[log_3(log 5)]$
=$\frac{1}{log3\times logx}$$\frac{d}{dx}[log x]$-0
=$\frac{1}{log3\times logx}\times \frac{1}{x}$
$\frac{dy}{dx}$=$\frac{1}{log3\times logx\times x}$
$2)y=log\left[e^{3x}.\frac{(3x-4)^\frac{2}{3}}{\sqrt[3]{\frac{a}{b}}}\right]$
Solution-
$=log\left[e^{3x}.(3x-4)^{\frac{2}{3}}\right]-log\left[(2x+5)^\frac{1}{3}\right]$
$=loge^{3x}+log(3x-4)^{\frac{2}{3}}-log(2x+5)^{\frac{1}{3}}$
$\therefore$ y=3x+$\frac{2}{3}[log(3x-4)]$-$\frac{1}{3}log(2x+5)$
differentiate w.r.t.x
$\frac{dy}{dx}$=$\frac{d}{dx}$$\{3x+[log (3x-4)-\frac{1}{3}log(2x+5)]\}$
$=x+\frac{1}{3x-4}$$\frac{d}{dx}(3x-4)$$-\frac{1}{3}$$\frac{1}{2x+5}$$\frac{d}{dx}(2x+5)$
=x+$\frac{3}{3x-4}-$$\frac{1}{3}$$\frac{2}{2x+5}$
3)$\frac{d}{dx}[log(logx)]$-$\frac{d}{dx}[log_3(log 5)]$
Solution-
=$\frac{1}{2}log{\frac{1-cos\frac{3x}{2}}{1+cos\frac{3x}{2}}}$
=$\frac{1}{2}$$\left[log(1-cos\frac{3x}{2})-log(1+cos\frac{3x}{2})\right]$
differentiate w.r.t x
$
\begin{math}
\frac{dy}{dx}=\frac{d}{dx}\{\frac{1}{2}\left[log(1-cos\frac{3x}{2})-log(1+cos\frac{3x}{2})\right]\}\\
=\frac{d}{dx}\{\frac{1}{2}\left[log(1-cos\frac{3x}{2})\}-\frac{d}{dx}\{log(1+cos\frac{3x}{2})\right]\}\vspace{1mm}\\
=\frac{1}{2}\Big\{\left[\frac{1}{log(1-cos\frac{3x}{2})}\frac{d}{dx}(1-cos\frac{3x}{2})\right]-\frac{1}{log(1+cos\frac{3x}{2})}\frac{d}{dx}(1+cos\frac{3x}{2})\Big\}\\
=\frac{1}{2}\Big\{\left[\frac{1}{log(1-cos\frac{3x}{2})}\times-sin\frac{3x}{2}\frac{d}{dx}\frac{3x}{2}\right]-\left[\frac{1}{log(1+cos\frac{3x}{2})}\times-sin\frac{3x}{2}\frac{d}{dx}\frac{3x}{2}\right]\Big\}\\
=\frac{1}{2}\Big\{\left[\frac{1}{log(1-cos\frac{3x}{2})}\times-sin\frac{3x}{2}\frac{3}{2}\right]-\left[\frac{1}{log(1+cos\frac{3x}{2})}\times-sin\frac{3x}{2}\frac{3}{2}\right]\Big\}\\
\hfill\break or\\
y=log\left[\sqrt{\frac{1-cos\frac{3x}{2}}{1+cos\frac{3x}{2}}}\right]\\
y=log\Big[\sqrt{\frac{2sin^{2}\big(\frac{3x}{4}\big)}{2cos^2\big(\frac{3x}{2}\big)}}\Big]\\
y=log\Big[tan\big(\frac{3x}{2}\big)\Big]\\
\end{math}$
Differentiate w.r.t x\\
\begin{math}
\frac{dy}{dx}=\frac{d}{dx}\Big\{log\left[tan\left(\frac{3x}{2}\right)\right]\Big\}\\
=\frac{1}{tan\frac{3x}{4}}\frac{d}{dx}\left[tan\frac{3x}{4}\right]\\
=cot\left(\frac{3x}{4}\right)sec^2(\frac{3x}{4})\frac{d}{dx}(\frac{3x}{4})\\
=cot\left(\frac{3x}{4}\right)sec^2(\frac{3x}{4})\frac{3}{4}\\
=\frac{cos\frac{3x}{2}}{sin\frac{3x}{2}}\times\frac{1}{cos^{2}{(\frac{3x}{2})}}\frac{3}{4}\\
=\frac{3}{2\left[2sin\frac{3x}{4}.cos\frac{3x}{4}\right]}\\
=\frac{3}{2sin\left(\frac{3x}{2}\right)}\\
\frac{dy}{dx}=\frac{3}{2}cosec\left(\frac{3x}{2}\right)\\
$4)y=log\left[\frac{x+\sqrt{x^2+a^2}}{\sqrt{x^2+a^2}-x}\right]$
$5)y=4^{log_2(sinx)}+9^{log_3(cosx)}$
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| differentiate Class 12th |
