Question 513 Marks
If $x^y= e ^{x-y}$, then show that $\frac{d y}{d x}=\frac{\log x}{(1+\log x)^2}$
View full question & answer→$e^{e^{x-y}}=\frac{x}{y}$
$e^{x+y}=\cos (x-y)$
$x e^y+y e^x=1$
$x^2 y^2-\tan ^{-1}\left(\sqrt{x^2+y^2}\right)=\cot ^{-1}\left(\sqrt{x^2+y^2}\right)$
$x^3+x^2 y+x y^2+y^3=81$
$x+\sqrt{x y}+y=1$
$\sin x^x$
$x^{\tan ^{-1} x}$
$\cot ^{-1}\left(\frac{4-x-2 x^2}{3 x+2}\right)$
$\tan ^{-1}\left(\frac{5-x}{6 x^2-5 x-3}\right)$
$\tan ^{-1}\left(\frac{a+b \tan x}{b-a \tan x}\right)$
$\cot ^{-1}\left(\frac{a^2-6 x^2}{5 a x}\right)$
$\tan ^{-1}\left(\frac{2^x}{1+2^{2 x+1}}\right)$
$\tan ^{-1}\left(\frac{2^{x+2}}{1-3\left(4^x\right)}\right)$
$\tan ^{-1}\left(\frac{2 \sqrt{x}}{1+3 x}\right)$
$\cot ^{-1}\left(\frac{1+35 x^2}{2 x}\right)$
$\operatorname{cosec}^{-1}\left(\frac{10}{6 \sin \left(2^x\right)-8 \cos \left(2^x\right)}\right)$
$\cos ^{-1}\left(\frac{3 \cos \left(e^x\right)+2 \sin \left(e^x\right)}{\sqrt{13}}\right)$
$\cos ^{-1}\left(\frac{3 \cos 3 x-4 \sin 3 x}{5}\right)$
$\cos ^{-1}\left(\sqrt{\frac{1+\cos x}{2}}\right)$
Differentiating w.r.t. $x$, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left(\frac{x}{2}\right)=\frac{1}{2} \frac{d}{d x}(x) \\ & =\frac{1}{2} \times 1=\frac{1}{2} .\end{aligned}$
$\sec ^{-1} x+\operatorname{cosec}^{-1} x=\frac{\pi}{2} \ldots[$ for $|x| \geq 1]$
Differentiating w.r.t. x, we get
$f^{\prime}(x)=\frac{d}{d x}\left(\sec ^{-1} x+\operatorname{cosec}^{-1} x\right)$
$=\frac{d}{d x}\left(\sec ^{-1} x\right)+\frac{d}{d x}\left(\operatorname{cosec}^{-1} x\right)$
$=\frac{1}{x \sqrt{x^2-1}}-\frac{1}{x \sqrt{x^2-1}}=0$.
Since, f'(x) = 0, f(x) is a constant function. Let f(x) = k. For any value of x, f(x) = k, where |x| > 1 Let x = 2. Then, f(2) = k ……(2)
From (1), $f(2)=\sec ^{-1}(2)+\operatorname{cosec}^{-1}(2)=\frac{\pi}{3}+\frac{\pi}{6}=\frac{\pi}{2}$
$\therefore k=\frac{\pi}{2} \quad \ldots$ [By (2)]
$\therefore f(x)=k=\frac{\pi}{2}$
Hence, $\sec ^{-1} x+\operatorname{cosec}^{-1} x=\frac{\pi}{2} . \quad \ldots$ [By (1)]
$\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$
Differentiating w.r.t. x, we get
$\begin{aligned} f^{\prime}(x) & =\frac{d}{d x}\left(\tan ^{-1} x+\cot ^{-1} x\right) \\ & =\frac{d}{d x}\left(\tan ^{-1} x\right)+\frac{d}{d x}\left(\cot ^{-1} x\right) \\ & =\frac{1}{1+x^2}-\frac{1}{1+x^2}=0\end{aligned}$
Since, f'(x) = 0, f(x) is a constant function. Let f(x) = k. For any value of x, f(x) = k Let x = 0. Then f(0) = k ….(2) From (1), f(0) = tan-1(0) + cot-1(0)
$=0+\frac{\pi}{2}=\frac{\pi}{2}$
$\therefore k=\frac{\pi}{2} \quad \ldots[$ By (2)]
$\therefore f(x)=k=\frac{\pi}{2}$
Hence, $\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2} . \quad \ldots$ [By (1)]
Now, $h(x)=\sqrt{4 f(x)+3 g(x)}$
$\begin{aligned} \therefore h^{\prime}(x) & =\frac{d}{d x}[\sqrt{4 f(x)+3 g(x)}] \\ & =\frac{1}{2 \sqrt{4 f(x)+3 g(x)}} \cdot \frac{d}{d x}[4 f(x)+3 g(x)] \\ & =\frac{1}{2 \sqrt{4 f(x)+3 g(x)}} \times\left[4 f^{\prime}(x)+3 g^{\prime}(x)\right] \\ \therefore h^{\prime}(1) & =\frac{1}{2 \sqrt{4 f(1)+3 g(1)}} \times\left[4 f^{\prime}(1)+3 g^{\prime}(1)\right] \\ & =\frac{1}{2 \sqrt{4 \times 4+3 \times 3}} \times[4 \times 3+3 \times 4] \ldots[\text { By }(1)] \\ & =\frac{1}{2 \sqrt{25}} \times 24 \\ & =\frac{1}{2 \times 5} \times 24=\frac{12}{5} .\end{aligned}$
$\frac{\left(x^2+2\right)^4}{\sqrt{x^2+5}}$
Differentiating w.r.t. x, we get
$\frac{d y}{d x}=\frac{d}{d x}\left[\frac{\left(x^2+2\right)^4}{\sqrt{x^2+5}}\right]$
$\begin{aligned} & =\frac{\sqrt{x^2+5} \cdot \frac{d}{d x}\left(x^2+2\right)^4-\left(x^2+2\right)^4 \cdot \frac{d}{d x}\left(\sqrt{x^2+5}\right)}{\left(\sqrt{x^2+5}\right)^2} \\ & \sqrt{x^2+5} \times 4\left(x^2+2\right)^3 \cdot \frac{d}{d x}\left(x^2+2\right)- \\ & =\frac{\left(x^2+2\right)^4 \times \frac{1}{2 \sqrt{x^2+5}} \cdot \frac{d}{d x}\left(x^2+5\right)}{x^2+5} \\ & =\frac{\sqrt{x^2+5} \times 4\left(x^2+2\right)^3 \cdot(2 x+0)-\frac{\left(x^2+2\right)^4}{2 \sqrt{x^2+5}} \times(2 x+0)}{x^2+5} \\ & =\frac{8 x\left(x^2+5\right)\left(x^2+2\right)^3-x\left(x^2+2\right)^4}{\left(x^2+5\right)^{\frac{3}{2}}} \\ & =\frac{x\left(x^2+2\right)^3\left[8\left(x^2+5\right)-\left(x^2+2\right)\right]}{\left(x^2+5\right)^{\frac{3}{2}}} \\ & =\frac{x\left(x^2+2\right)^3\left(8 x^2+40-x^2-2\right)}{\left(x^2+5\right)^{\frac{3}{2}}} \\ & =\frac{x\left(x^2+2\right)^3\left(7 x^2+38\right)}{\left(x^2+5\right)^{\frac{3}{2}}} . \\ & \end{aligned}$
$y=(25)^{\log _5(\sec x)}-(16)^{\log _4(\tan x)}$
Differentiating w.r.t. x, we get
$\frac{d y}{d x}=\frac{d}{d x}(1)=0$
$\log \left[\frac{a^{\cos x}}{\left(x^2-3\right)^3 \log x}\right]$
Differentiating w.r.t. $x$, we get
$\frac{d y}{d x}=\frac{d}{d x}\left[(\cos x)(\log a)-3 \log \left(x^2-3\right)-\log (\log x)\right]$
$\begin{aligned} & =(\log a) \cdot \frac{d}{d x}(\cos x)-3 \frac{d}{d x}\left[\log \left(x^2-3\right)\right]-\frac{d}{d x}[\log (\log x)] \\ & =(\log a)(-\sin x)-3 \times \frac{1}{x^2-3} \cdot \frac{d}{d x}\left(x^2-3\right)-\frac{1}{\log x} \cdot \frac{d}{d x}(\log x) \\ & =-(\sin x)(\log a)-\frac{3}{x^2-3} \times(2 x-0)-\frac{1}{\log x} \times \frac{1}{x} \\ & =-(\sin x)(\log a)-\frac{6 x}{x^2-3}-\frac{1}{x \log x} .\end{aligned}$
$\log \left[\frac{e^{x^2}(5-4 x)^{\frac{3}{2}}}{\sqrt[3]{7-6 x}}\right]$
Using $ \log (A \cdot B)=\log A+\log B $
$\begin{aligned} & y=\log e^{x^2}+\log \left(\frac{(5-4 x)^{\frac{3}{2}}}{\sqrt[3]{7-6 x}}\right) \\ & =\log e^{x^2}+\log (5-4 x)^{\frac{3}{2}}-\log (\sqrt[3]{7-6 x}) \\ & =x^2 \log e+\frac{3}{2} \log (5-4 x)-\log (7-6 x)^{\frac{1}{3}} \\ & =x^2+\frac{3}{2} \log (5-4 x)-\frac{1}{3} \log (7-6 x)\end{aligned}$
Now, Differentiating w.r.t. $x$, we get
$\begin{aligned} & \frac{ dy }{ dx }=\frac{ d }{ dx } x^2+\frac{3}{2} \frac{ d }{ dx } \log (5-4 x)-\frac{1}{3} \frac{ d }{ dx } \log (7-6 x) \\ & =2 x+\frac{3}{2} \frac{1}{5-4 x}(-4)-\frac{1}{3} \frac{1}{(7-6 x)} x(-6) \\ & =2 x-\frac{6}{(5-4 x)}+\frac{2}{(7-6 x)} \\ & 2 x-\frac{6}{5-4 x}+\frac{2}{7-6 x} .\end{aligned}$
$\log \left[4^{2 x}\left(\frac{x^2+5}{\sqrt{2 x^3-4}}\right)^{\frac{3}{2}}\right]$
Differentiating w.r.t. $x$, we get
$\begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}\left[2 x \log 4+\frac{3}{2} \log \left(x^2+5\right)-\frac{3}{4} \log \left(2 x^3-4\right)\right] \\ & =(2 \log 4) \frac{d}{d x}(x)+\frac{3}{2} \frac{d}{d x}\left[\log \left(x^2+5\right)\right]-\frac{3}{4} \frac{d}{d x}\left[\log \left(2 x^3-4\right)\right] \\ & =(2 \log 4) \times 1+\frac{3}{2} \times \frac{1}{x^2+5} \cdot \frac{d}{d x}\left(x^2+5\right)- \\ & =2 \log 4+\frac{3}{2\left(x^2+5\right)} \times(2 x+0)-\frac{3}{4\left(2 x^3-4\right)} \times \frac{1}{2 x^3-4} \cdot \frac{d}{d x}\left(2 x^3-4\right) \\ & =2 \log 4+\frac{3 x}{x^2+5}-\frac{\left.9 x^2-0\right)}{2\left(2 x^3-4\right)} .\end{aligned}$
$\log \left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right)$
$=\log \left(\frac{1-\sin x}{\cos x}\right)$
$\begin{aligned} & =\log \left(\frac{1}{\cos x}-\frac{\sin x}{\cos x}\right) \\ & =\log (\sec x-\tan x)\end{aligned}$
Differentiating w.r.t. $x$, we get
$\begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}[\log (\sec x-\tan x)] \\ & =\frac{1}{\sec x-\tan x} \cdot \frac{d}{d x}(\sec x-\tan x) \\ & =\frac{1}{\sec x-\tan x} \times\left(\sec x \tan x-\sec ^2 x\right) \\ & =\frac{-\sec x(\sec x-\tan x)}{\sec x-\tan x} \\ & =-\sec x \\ & \end{aligned}$
$\log \left(\sqrt{\left.\frac{1+\cos \left(\frac{5 x}{2}\right)}{1-\cos \left(\frac{5 x}{2}\right)}\right)}\right.$
$\log a^b=b \log a$
$\begin{aligned} & y=\log \left(\sqrt{1+\cos \frac{5 x}{2}}\right)-\log \left(\sqrt{1-\cos \left(\frac{5 x}{2}\right)}\right) \\ & y=\log \left(1+\cos \left(\frac{x}{2}\right)^{\frac{1}{2}}-\log \left(1-\cos \left(\frac{5 x}{2}\right)\right)^{\frac{1}{2}}\right. \\ & y=\frac{1}{2} \log \left[1+\cos \left(\frac{5 x}{2}\right)\right]-\frac{1}{2} \log \left[\left(1-\cos \left(\frac{5 x}{2}\right)\right]\right.\end{aligned}$
Differentiating w.r.t. x
$\begin{aligned} & \frac{d y}{d x}=\frac{1}{2} \frac{1}{1+\cos \left(\frac{5 x}{2}\right)} \frac{d}{d x}\left(1+\cos \frac{5 x}{2}\right)-\frac{1}{2} \times \frac{1}{1-\cos \left(\frac{5 x}{2}\right)} \frac{d}{d x}\left(1-\cos \frac{5 x}{2}\right) \\ & =\frac{1}{2\left(1+\cos \left(\frac{5 x}{2}\right)\right)}\left(-\sin \left(\frac{5 x}{2}\right) \cdot \frac{5}{2}-\frac{1}{2\left(1-\cos \left(\frac{5 x}{2}\right)\right)}\left(\sin \left(\frac{5 x}{2}\right) \cdot \frac{5}{2}\right.\right. \\ & =\frac{-5 \sin \left(\frac{5 x}{2}\right)}{4\left(1+\cos \left(\frac{5 x}{2}\right)\right)}-\frac{5 \sin \left(\frac{5 x}{2}\right)}{4\left(1-\cos \left(\frac{5 x}{2}\right)\right)} \\ & =\frac{-5}{4} \sin \left(\frac{5 x}{2}\right)\left[\frac{1}{1+\cos ^2\left(\frac{5 x}{2}\right)}+\frac{1}{1-\cos ^2\left(\frac{5 x}{2}\right)}\right] \\ & =\frac{\frac{-5}{2} \sin \left(\frac{5 x}{2}\right)\left[1-\cos ^2\left(\frac{5 x}{2}\right)+1+\cos \frac{5 x}{2}\right]}{\left[1-\cos ^2\left(\frac{5 x}{2}\right)\right]} \\ & =\frac{-5}{4} \sin \left(\frac{5 x}{2}\right) \times \frac{2}{\sin ^2\left(\frac{5 x}{2}\right)} \ldots\left[\because 1-\cos ^2 x=\sin ^2 x\right] \\ & =\frac{-5}{4} \frac{1}{\sin \left(\frac{5 x}{2}\right)} \\ & -\frac{5}{2} \times \cos \cos \left(\frac{5 x}{2}\right)\end{aligned}$
$\log \left(\sqrt{\frac{1-\cos 3 x}{1+\cos 3 x}}\right)$
Differentiating w.r.t. $x$, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[\log \tan \left(\frac{3 x}{2}\right)\right] \\ & =\frac{1}{\tan \left(\frac{3 x}{2}\right)} \times \frac{d}{d x}\left[\tan \left(\frac{3 x}{2}\right)\right] \\ & =\frac{1}{\tan \left(\frac{3 x}{2}\right)} \times \sec ^2\left(\frac{3 x}{2}\right) \cdot \frac{d}{d x}\left(\frac{3 x}{2}\right)\end{aligned}$
$\begin{aligned} & =\frac{\cos \left(\frac{3 x}{2}\right)}{\sin \left(\frac{3 x}{2}\right)} \times \frac{1}{\cos ^2\left(\frac{3 x}{2}\right)} \times \frac{3}{2} \times 1 \\ & =3 \times \frac{1}{2 \sin \left(\frac{3 x}{2}\right) \cos \left(\frac{3 x}{2}\right)} \\ & =3 \times \frac{1}{\sin 3 x}=3 \operatorname{cosec} 3 x .\end{aligned}$
$\log \left[\tan ^3 x \cdot \sin ^4 x \cdot\left(x^2+7\right)^7\right]$
Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[3 \log \tan x+4 \log \sin x+7 \log \left(x^2+7\right)\right] \\ & =3 \frac{d}{d x}(\log \tan x)+4 \frac{d}{d x}(\log \sin x)+7 \frac{d}{d x}\left[\log \left(x^2+7\right)\right] \\ & =3 \times \frac{1}{\tan x} \cdot \frac{d}{d x}(\tan x)+4 \times \frac{1}{\sin x} \cdot \frac{d}{d x}(\sin x)+ \\ & =3 \times \frac{1}{\tan x} \cdot \sec 2 x+4 \times \frac{1}{\sin x} \cdot \cos x+7 \times \frac{1}{x^2+7} \cdot(2 x+0) \\ & =3 \times \frac{\cos x}{\sin x} \times \frac{1}{\cos ^2 x}+4 \cot x+\frac{1}{x^2+7} \cdot \frac{d}{d x}\left(x^2+7\right) \\ & =\frac{6}{2 \sin x \cos x}+4 \cot x+\frac{14 x}{x^2+7} \\ & =\frac{6}{\sin 2 x}+4 \cot ^2 x+\frac{14 x}{x^2+7} \\ & =6 \operatorname{cosec} 2 x+4 \cot x+\frac{14 x}{x^2+7}\end{aligned}$