Question 512 Marks
Diffrentiate the following w. r. t. x
View full question & answer→$\cos ^{-1}\left(3 x-4 x^3\right)$
$\cos ^{-1}\left(3 x-4 x^3\right)$
$\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)$
$\sin ^{-1}\left(\frac{1-x^2}{1+x^2}\right)$
$\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)$
$\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)$
$\operatorname{cosec}^{-1}\left(\frac{1}{\cos \left(5^x\right)}\right)$
Differentiating w.r.t. $x$, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left(\frac{\pi}{2}-5^x\right) \\ & =\frac{d}{d x}\left(\frac{\pi}{2}\right)-\frac{d}{d x}\left(5^x\right) \\ & =0-5^x \cdot \log 5 \\ & =-5^x \cdot \log 5\end{aligned}$
y = x logx
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}(x \log x) \\ & =x \frac{d}{d x}(\log x)+(\log x) \cdot \frac{d}{d x}(x) \\ & =x \times \frac{1}{x}+(\log x) \times 1 \\ & =1+\log x\end{aligned}$
The derivative of inverse function of $y=f(x)$ is given by
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}(x \log x) \\ & =x \frac{d}{d x}(\log x)+(\log x) \cdot \frac{d}{d x}(x) \\ & =x \times \frac{1}{x}+(\log x) \times 1 \\ & =1+\log x\end{aligned}$
The derivative of inverse function of $y=f(x)$ is
given by
$\frac{d x}{d y}=\frac{1}{\left(\frac{d y}{d x}\right)}=\frac{1}{1+\log x}$
$y=x^2+\log x$
Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left(x^2+\log x\right) \\ & =\frac{d}{d x}\left(x^2\right)+\frac{d}{d x}(\log x) \\ & =2 x+\frac{1}{x}=\frac{2 x^2+1}{x}\end{aligned}$
The derivative of inverse function of $y=f(x)$ is
given by
$\frac{d x}{d y}=\frac{1}{\left(\frac{d y}{d x}\right)}=\frac{1}{\left(\frac{2 x^2+1}{x}\right)}=\frac{x}{2 x^2+1}$
y = x·7x
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left(x \cdot 7^x\right) \\ & =x \frac{d}{d x}\left(7^x\right)+7^x \frac{d}{d x}(x) \\ & =x \cdot 7^x \log 7+7^x \times 1 \\ & =7^x(x \log 7+1)\end{aligned}$
The derivative of inverse function of $y=f(x)$ is
given by
$\frac{d x}{d y}=\frac{1}{\left(\frac{d y}{d x}\right)}=\frac{1}{7^x(x \log 7+1)}$
y = x cos x
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}(x \cos x) \\ & =x \frac{d}{d x}(\cos x)+\cos x \frac{d}{d x}(x) \\ & =x(-\sin x)+\cos x \times 1 \\ & =\cos x-x \sin x\end{aligned}$
The derivative of inverse function of $y=f(x)$ is
given by
$\frac{d x}{d y}=\frac{1}{\left(\frac{d y}{d x}\right)}=\frac{1}{\cos x-x \sin x}$
$y=x^2 \cdot e^x$
Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left(x^2 \cdot e^x\right) \\ & =x^2 \frac{d}{d x}\left(e^x\right)+e^x \frac{d}{d x}\left(x^2\right) \\ & =x^2 \cdot e^x+e^x \times 2 x \\ & =x e^x(x+2)\end{aligned}$
The derivative of inverse function of $y=f(x)$ is
given by
$\frac{d x}{d y}=\frac{1}{\left(\frac{d y}{d x}\right)}=\frac{1}{x e^x(x+2)}$
$y=\log _2\left(\frac{x}{2}\right)$
We have to find the inverse function of y = f(x), i.e. x in terms of y. From (1),
$\begin{aligned} & \frac{x}{2}=2^y \therefore x=2 \cdot 2^y=2^{y+1} \\ & \therefore x=f^{-1}(y)=2^{y+1}\end{aligned}$
$\begin{aligned} \therefore \frac{d x}{d y} & =\frac{d}{d y}\left(2^{y+1}\right) \\ & =2^{y+1} \cdot \log 2 \cdot \frac{d}{d y}(y+1) \\ & =2^{y+1} \cdot \log 2 \cdot(1+0) \\ & =2^{y+1} \cdot \log 2=2^{\log _2\left(\frac{x}{2}\right)+1} \cdot \log 2 \ldots \text { [Вy (1)] }\end{aligned}$
$\begin{aligned} & =2^{\log _2\left(\frac{x}{2}\right)+\log _2 2} \cdot \log 2 \\ & =2^{\log _2\left(\frac{x}{2} \times 2\right)} \cdot \log 2=2^{\log _2 x} \cdot \log 2 \\ & =x \log 2 \quad \ldots\left[\because a^{\log _5 x}=x\right]\end{aligned}$
$\therefore \frac{d y}{d x}=\frac{1}{\left(\frac{d x}{d y}\right)}=\frac{1}{x \log 2}$
$y=e^{2 x-3}$
We have to find the inverse function of y = f(x), i.e. x in terms of y. From (1), 2x – 3 = log y ∴ 2x = log y + 3
$\begin{aligned} \therefore x & =f^{-1}(y)=\frac{1}{2}(\log y+3) \\ \therefore \frac{d x}{d y} & =\frac{1}{2} \frac{d}{d y}(\log y+3) \\ & =\frac{1}{2}\left(\frac{1}{y}+0\right)=\frac{1}{2 y}\end{aligned}$
$=\frac{1}{2 e^{2 x-3}}$
... [By (1)]
$\therefore \frac{d y}{d x}=\frac{1}{\left(\frac{d x}{d y}\right)}=\frac{1}{\left(\frac{1}{2 e^{2 x-3}}\right)}=2 e^{2 x-3}$
$y=e^x-3$
We have to find the inverse function of y = f(x), i.e. x in terms of y. From (1), ex = y + 3 ∴ x = log(y + 3) ∴ x = f-1(y) = log(y + 3)
$\begin{aligned} \therefore \frac{d x}{d y} & =\frac{d}{d y}[\log (y+3)] \\ & =\frac{1}{y+3} \cdot \frac{d}{d y}(y+3) \\ & =\frac{1}{y+3} \cdot(1+0)=\frac{1}{y+3} \\ & =\frac{1}{e^x-3+3} \\ & =\frac{1}{e^x}\end{aligned}$
... [By (1)]
$\therefore \frac{d y}{d x}=\frac{1}{\left(\frac{d x}{d y}\right)}=\frac{1}{\left(\frac{1}{e^x}\right)}=e^x$.
$\begin{aligned} & 2 x=y-3 \quad \therefore x=\frac{y-3}{2} \\ & \therefore x=f^{-1}(y)=\frac{y-3}{2}\end{aligned}$
$\therefore \frac{d x}{d y}=\frac{1}{2} \frac{d}{d y}(y-3)$
$\begin{aligned} \quad & \frac{1}{2}(1-0)=\frac{1}{2} \\ \therefore \frac{d y}{d x} & =\frac{1}{\left(\frac{d x}{d y}\right)}=\frac{1}{\left(\frac{1}{2}\right)}=2\end{aligned}$
y = 2x + 3 ….(1) We have to find the inverse function of y = f(x), i.e. x in terms of y. From (1),
$\begin{aligned} & 2 x-1=e^y \quad \therefore 2 x=e^y+1 \\ & \therefore x=f^{-1}(y)=\frac{1}{2}\left(e^y+1\right) \\ & \therefore \frac{d x}{d y}=\frac{1}{2} \frac{d}{d y}\left(e^y+1\right) \\ & \quad=\frac{1}{2}\left(e^y+0\right)=\frac{1}{2} e^y\end{aligned}$
$=\frac{1}{2} e^{\log (2 x-1)} \quad \ldots$ [By (1)]
$=\frac{1}{2}(2 x-1) \quad \ldots\left[\because e^{\log x}=x\right]$
$\therefore \frac{d y}{d x}=\frac{1}{\left(\frac{d x}{d y}\right)}=\frac{2}{2 x-1}$
$y=\sqrt[3]{x-2}$
We have to find the inverse function of y = f(x), i.e. x in terms of y. From (1),
$\begin{aligned} & y^3=x-2 \quad \therefore x=y^3+2 \\ & \therefore x=f^{-1}(y)=y^3+2 \\ & \therefore \frac{d x}{d y}=\frac{d}{d y}\left(y^3+2\right) \\ & \quad=3 y^2+0=3 y^2\end{aligned}$
$=3(\sqrt[3]{(x-2)})^2 \quad \ldots[$ By (1)]
$=3(x-2)^3=3 \cdot\left(\sqrt[3]{(x-2)^2}\right)$
$\therefore \frac{d y}{d x}=\frac{1}{\left(\frac{d x}{d Y}\right)}=\frac{1}{\left.3 \sqrt[3]{(x-2)^2}\right)}, x>2$
$y=\sqrt{2-\sqrt{x}}$
We have to find the inverse function of y = f(x), i.e. x in terms of y. From (1),
$\begin{aligned} & y^2=2-\sqrt{x} \quad \therefore \sqrt{x}=2-y^2 \\ & \therefore x=\left(2-y^2\right)^2 \\ & \therefore x=f^{-1}(y)=\left(2-y^2\right)^2 \\ & \therefore \frac{d x}{d y}=\frac{d}{d y}\left(2-y^2\right)^2 \\ & \quad=2\left(2-y^2\right) \cdot \frac{d}{d y}\left(2-y^2\right) \\ & \quad=2\left(2-y^2\right) \cdot(0-2 y) \\ & \quad=-4 y\left(2-y^2\right)\end{aligned}$
$\begin{aligned} & =-4 \sqrt{2-\sqrt{x}}(2-2+\sqrt{x}) \quad \ldots \text { [By (1)] } \\ & =-4 \sqrt{x} \sqrt{2-\sqrt{x}}\end{aligned}$
$\therefore \frac{d y}{d x}=\frac{1}{\left(\frac{d x}{d y}\right)}=-\frac{1}{4 \sqrt{x} \sqrt{2-\sqrt{x}}}$
$y=\sqrt{x}$
We have to find the inverse function of y = f(x), i.e. x in terms of y. From (1),
$\begin{aligned} & y ^2= x \therefore x = y ^2 \\ & \therefore x=f^{-1}(y)=y^2 \\ & \therefore \frac{d x}{d y}=\frac{d}{d y}\left(y^2\right)=2 y\end{aligned}$
$\begin{aligned} & =2 \sqrt{x} \\ \therefore \frac{d y}{d x} & =\frac{1}{\left(\frac{d x}{d y}\right)}=\frac{1}{2 \sqrt{x}} .\end{aligned}$
$\ldots$ [By (1)]
The derivative of f[g(x)] w.r.t. x in terms of and g is f'[g(x)]∙g'(x)
$\therefore \frac{d}{d x}\{ f [ g ( x )]\}=2 e ^{2 x }+6 e ^{ x }$ and $\frac{d}{d x}\{ f [ g ( x )]\}_{ x }=0=8$
The derivative of g[f(x)] w.r.t. x in terms of f and g is g’f(x)]∙f'(x).
$\begin{aligned} & \therefore \frac{d}{d x}\left\{ g [( f ( x )]\}=2 xe ^{ x 2+5} \text { and }\right. \\ & \frac{d}{d x}\left\{ g [( f ( x )]\}_{ x }=-1\right. \\ & =-2 e ^6\end{aligned}$
$\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^5$
Differentiating w.r.t. x, we get
$\frac{d y}{d x}=\frac{d}{d x}\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^5$
$\begin{aligned} & =5\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^4 \cdot \frac{d}{d x}\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right) \\ & =5\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^4 \cdot\left[\frac{d}{d x}(3 x-5)^{\frac{1}{2}}-\frac{d}{d x}(3 x-5)^{-\frac{1}{2}}\right] \\ & =5\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^4 \times \\ & {\left[\frac{1}{2}(3 x-5)^{-\frac{1}{2}} \cdot \frac{d}{d x}(3 x-5)-\left(-\frac{1}{2}\right)(3 x-5)^{-\frac{3}{2}} \cdot \frac{d}{d x}(3 x-5)\right]} \\ & =5\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^4 \times \\ & {\left[\frac{1}{2 \sqrt{3 x-5}} \cdot(3 \times 1-0)+\frac{1}{2(3 x-5)^{\frac{3}{2}}} \cdot(3 \times 1-0)\right]} \\ & =5\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^4 \cdot\left[\frac{3}{2 \sqrt{3 x-5}}+\frac{3}{2(3 x-5)^{\frac{3}{2}}}\right] \\ & =\frac{15}{2}\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^4 \cdot\left[\frac{3 x-5+1}{(3 x-5)^{\frac{3}{2}}}\right] \\ & =\frac{15(3 x-4)}{2(3 x-5)^{\frac{3}{2}}}\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^4 \text {. } \\ & \end{aligned}$
$\frac{3}{5 \sqrt[3]{\left(2 x^2-7 x-5\right)^5}}$
Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{3}{5} \frac{d}{d x}\left(2 x^2-7 x-5\right)^{-\frac{5}{3}} \\ & =\frac{3}{5} \times\left(-\frac{5}{3}\right)\left(2 x^2-7 x-5\right)^{-\frac{5}{3}-1} \cdot \frac{d}{d x}\left(2 x^2-7 x-5\right) \\ & =-\left(2 x^2-7 x-5\right)^{-\frac{8}{3}} \cdot(2 \times 2 x-7 \times 1-0) \\ & =-\frac{4 x-7}{\left(2 x^2-7 x-5\right)^{\frac{8}{3}}} \cdot\end{aligned}$
$\sqrt{x^2+\sqrt{x^2+1}}$
Differentiating w.r.t. x, we get
$\frac{d y}{d x}=\frac{d}{d x}\left(x^2+\sqrt{x^2+1}\right)^{\frac{1}{2}}$
$\begin{aligned} & =\frac{1}{2}\left(x^2+\sqrt{x^2+1}\right)^{-\frac{1}{2}} \cdot \frac{d}{d x}\left(x^2+\sqrt{x^2+1}\right) \\ & =\frac{1}{2 \sqrt{x^2+\sqrt{x^2+1}}} \cdot\left[\frac{d}{d x}\left(x^2\right)+\frac{d}{d x}\left(\sqrt{x^2+1}\right)\right] \\ & =\frac{1}{2 \sqrt{x^2+\sqrt{x^2+1}}} \cdot\left[2 x+\frac{1}{2 \sqrt{x^2+1}} \cdot \frac{d}{d x}\left(x^2+1\right)\right] \\ & =\frac{1}{2 \sqrt{x^2+\sqrt{x^2+1}}} \cdot\left[2 x+\frac{1}{2 \sqrt{x^2+1}}(2 x+0)\right] \\ & =\frac{1}{2 \sqrt{x^2+\sqrt{x^2+1}}} \cdot\left[2 x+\frac{x}{\sqrt{x^2+1}}\right] \\ & =\frac{1}{2 \sqrt{x^2+\sqrt{x^2+1}}} \cdot\left[\frac{2 x \sqrt{x^2+1}+x}{\sqrt{x^2+1}}\right] \\ & 2 \sqrt{x\left(2 \sqrt{x^2+1} \cdot \sqrt{x^2+1}+1\right)} \cdot \sqrt{x^2+1} \cdot\end{aligned}$
$\sqrt{x^2+4 x-7}$
Differentiating w.r.t. $x$, we get
$\begin{aligned} & \frac{ dy }{ dx }=\frac{1}{2 \sqrt{x^2+4 x-7}} \cdot \frac{ d }{ dx }\left(x^2+4 x-7\right) \\ & =\frac{1}{2 \sqrt{x^2+4 x-7}}\left(\frac{ d }{ dx } x^2+\frac{ d }{ dx } 4 x-\frac{ d }{ dx } 7\right) \\ & =\frac{1}{2 \sqrt{x^2+4 x-7}} \cdot(2 x+4-0) \\ & =\frac{2(x+2)}{2 \sqrt{x^2+4 x-7}} \\ & =\frac{(x+2)}{\sqrt{x^2+4 x-7}} .\end{aligned}$
$\left(2 x^{\frac{3}{2}}-3 x^{\frac{4}{3}}-5\right)^{\frac{5}{2}}$
Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left(2 x^{\frac{3}{2}}-3 x^{\frac{4}{3}}-5\right)^{\frac{5}{2}} \\ & =\frac{5}{2}\left(2 x^{\frac{3}{2}}-3 x^{\frac{4}{3}}-5\right)^{\frac{5}{2}-1} \times \frac{d}{d x}\left(2 x^{\frac{3}{2}}-3 x^{\frac{4}{3}}-5\right) \\ & =\frac{5}{2}\left(2 x^{\frac{3}{2}}-3 x^{\frac{4}{3}}-5\right)^{\frac{3}{2}} \times\left(2 \times \frac{3}{2} x^{\frac{3}{2}-1}-3 \times \frac{4}{3} x^{\frac{4}{3}-1}-0\right) \\ & =\frac{5}{2}\left(2 x^{\frac{3}{2}}-3 x^{\frac{4}{3}}-5\right)^{\frac{3}{2}}\left(3 x^{\frac{1}{2}}-4 x^{\frac{1}{3}}\right) \\ & =\frac{5}{2}(3 \sqrt{x}-4 \sqrt[3]{x})\left(2 x^{\frac{3}{2}}-3 x^{\frac{4}{3}}-5\right)^{\frac{3}{2}}\end{aligned}$
$\left(x^3-2 x-1\right)^5$
$\therefore \frac{d y}{d u}=\frac{d}{d u}\left(u^5\right)=5 u^4$
$\begin{aligned} & =5\left(x^3-2 x-1\right)^4 \\ \text { and } \frac{d u}{d x} & =\frac{d}{d x}\left(x^3-2 x-1\right) \\ & =3 x^2-2 \times 1-0=3 x^2-2 \\ \therefore \frac{d y}{d x} & =\frac{d y}{d u} \times \frac{d u}{d x} \\ & =5\left(x^3-2 x-1\right)^4\left(3 x^2-2\right) \\ & =5\left(3 x^2-2\right)\left(x^3-2 x-1\right)^4 .\end{aligned}$
Method 2: Let $y=\left(x^3-2 x-1\right)^5$ Differentiating w.r.t. $x$, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left(x^3-2 x-1\right)^5 \\ & =5\left(x^3-2 x-1\right)^4 \times \frac{d}{d x}\left(x^3-2 x-1\right) \\ & =5\left(x^3-2 x-1\right)^4 \times\left(3 x^2-2 \times 1-0\right) \\ & =5\left(3 x^2-2\right)\left(x^3-2 x-1\right)^4\end{aligned}$
$\begin{array}{rlr}\therefore\left[\frac{d y}{d x}\right]_{x=2} & =f^{\prime}[g(2)] \cdot g^{\prime}(2) & \\ & =f^{\prime}(3) \cdot g^{\prime}(2) & \ldots[\because g(2)=3] \\ & =-1 \times 5 & \ldots \text { (Given) } \\ & =-5 .\end{array}$
$\sin ^2 x^2-\cos ^2 x^2$
Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[\sin ^2 x^2-\cos ^2 x^2\right] \\ & =\frac{d}{d x}\left(\sin x^2\right)^2-\frac{d}{d x}\left(\cos x^2\right)^2 \\ & =2 \sin x^2 \cdot \frac{d}{d x}\left(\sin x^2\right)-2 \cos x^2 \cdot \frac{d}{d x}\left(\cos x^2\right) \\ & =2 \sin x^2 \cdot \cos x^2 \cdot \frac{d}{d x}\left(x^2\right)-2 \cos x^2 \cdot\left(-\sin x^2\right) \cdot \frac{d}{d x}\left(x^2\right) \\ & =2 \sin x^2 \cdot \cos x^2 \times 2 x+2 \sin x^2 \cdot \cos x^2 \times 2 x \\ & =4 x\left(2 \sin x^2 \cdot \cos x^2\right) \\ & =4 x \sin \left(2 x^2\right) .\end{aligned}$
$[\log \{\log (\log x)\}]^2$
Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}[\log \{\log (\log x)\}]^2 \\ & =2 \cdot \log \{\log (\log x)\} \times \frac{d}{d x}[\log \{\log (\log x)\}]\end{aligned}$
$\begin{aligned} & =2 \cdot \log \{\log (\log x)\} \times \frac{1}{\log (\log x)} \cdot \frac{d}{d x}[\log (\log x)] \\ & =2 \cdot \log \{\log (\log x)\} \times \frac{1}{\log (\log x)} \times \frac{1}{\log x} \times \frac{d}{d x}(\log x) \\ & =2 \cdot \log \{\log (\log x)\} \times \frac{1}{\log (\log x)} \times \frac{1}{\log x} \times \frac{1}{x} \\ & =2 \cdot\left[\frac{\log \{\log (\log x)\}}{x \cdot \log x \cdot \log (\log x)}\right] .\end{aligned}$
$\log _{e^2}(\log x)$
$=\frac{\log (\log x)}{2 \log e}=\frac{\log (\log x)}{2} \quad \ldots[\because \log e=1]$
Differentiating w.r.t. $x$, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{1}{2} \frac{d}{d x}[\log (\log x)] \\ & =\frac{1}{2} \times \frac{1}{\log x} \cdot \frac{d}{d x}(\log x) \\ = & \frac{1}{2 \log x} \times \frac{1}{x}=\frac{1}{2 x \log x}\end{aligned}$
$\log \left[\sec \left( e ^{\times 2}\right)\right]$
Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[\log \left(\sec e^{x^2}\right)\right] \\ & =\frac{1}{\sec \left(e^{x^2}\right)} \cdot \frac{d}{d x}\left[\sec \left(e^{x^2}\right)\right] \\ & =\frac{1}{\sec \left(e^{x^2}\right)} \cdot \sec \left(e^{x^2}\right) \tan \left(e^{x^2}\right) \cdot \frac{d}{d x}\left(e^{x^2}\right) \\ & =\tan \left(e^{x^2}\right) \cdot e^{x^2} \cdot \frac{d}{d x}\left(x^2\right) \\ & =\tan \left(e^{x^2}\right) \cdot e^{x^2} \cdot 2 x \\ & =2 x \cdot e^{x^2} \tan \left(e^{x^2}\right) .\end{aligned}$
$\sin \sqrt{\sin \sqrt{x}}$
Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}(\sin \sqrt{\sin \sqrt{x}}) \\ & =\cos \sqrt{\sin \sqrt{x}} \cdot \frac{d}{d x}(\sqrt{\sin \sqrt{x}}) \\ & =\cos \sqrt{\sin \sqrt{x}} \times \frac{1}{2 \sqrt{\sin \sqrt{x}}} \cdot \frac{d}{d x}(\sin \sqrt{x}) \\ & =\frac{\cos \sqrt{\sin \sqrt{x}}}{2 \sqrt{\sin \sqrt{x}}} \times \cos \sqrt{x} \cdot \frac{d}{d x}(\sqrt{x})\end{aligned}$
$\begin{aligned} & =\frac{\cos \sqrt{\sin \sqrt{x}} \cdot \cos \sqrt{x}}{2 \sqrt{\sin \sqrt{x}}} \times \frac{1}{2 \sqrt{x}} \\ & =\frac{\cos \sqrt{\sin \sqrt{x}} \cdot \cos \sqrt{x}}{4 \sqrt{x} \cdot \sqrt{\sin \sqrt{x}}} .\end{aligned}$
$e^{\log [(\log x) 2-\log x 2]}$
Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[(\log x)^2-2 \log x\right] \\ & =\frac{d}{d x}(\log x)^2-2 \frac{d}{d x}(\log x) \\ & =2 \log x \cdot \frac{d}{d x}(\log x)-2 \times \frac{1}{x} \\ & =2 \log x \times \frac{1}{x}-\frac{2}{x} \\ & =\frac{2 \log x}{x}-\frac{2}{x} .\end{aligned}$
$\sec \left[\tan \left(x^4+4\right)\right]$
Differentiating w.r.t. x, we get
$\begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}\left\{\sec \left[\tan \left(x^4+4\right)\right]\right\} \\ & =\sec \left[\tan \left(x^4+4\right)\right] \cdot \tan \left[\tan \left(x^4+4\right)\right] \cdot \frac{d}{d x}\left[\tan \left(x^4+4\right)\right] \\ & =\sec \left[\tan \left(x^4+4\right)\right] \cdot \tan \left[\tan \left(x^4+4\right)\right] \cdot \sec ^2\left(x^4+4\right) \cdot \frac{d}{d x}\left(x^4+4\right) \\ & =\sec \left[\tan \left(x^4+4\right)\right] \cdot \tan \left[\tan \left(x^4+4\right)\right] \cdot \sec ^2\left(x^4+4\right)\left(4 x^3+0\right) \\ & =4 x^3 \sec ^2\left(x^4+4\right) \cdot \sec \left[\tan \left(x^4+4\right)\right] \cdot \tan \left[\tan \left(x^4+4\right)\right]\end{aligned}$
$\tan [\cos (\sin x)]$
$\frac{d y}{d x}=\frac{d}{d x}\{\tan [\cos (\sin x)]\}$
$\begin{aligned} & =\sec ^2[\cos (\sin x)] \cdot \frac{d}{d x}[\cos (\sin x)] \\ & =\sec ^2[\cos (\sin x)] \cdot[-\sin (\sin x)] \cdot \frac{d}{d x}(\sin x) \\ & =-\sec ^2[\cos (\sin x)] \cdot \sin (\sin x) \cdot \cos x .\end{aligned}$
$\cos ^2\left[\log \left(x^2+7\right)\right]$
Differentiating w.r.t. x, we get
$\begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}\left\{\cos \left[\log \left(x^2+7\right)\right]\right\}^2 \\ & =2 \cos \left[\log \left(x^2+7\right)\right] \cdot \frac{d}{d x}\left\{\cos \left[\log \left(x^2+7\right)\right]\right\} \\ & =2 \cos \left[\log \left(x^2+7\right)\right] \cdot\left\{-\sin \left[\log \left(x^2+7\right)\right]\right\} \cdot \frac{d}{d x}\left[\log \left(x^2+7\right)\right] \\ & =-2 \sin \left[\log \left(x^2+7\right)\right] \cdot \cos \left[\log \left(x^2+7\right)\right] \times \frac{1}{x^2+7} \cdot \frac{d}{d x}\left(x^2+7\right) \\ & =-\sin \left[2 \log \left(x^2+7\right)\right] \times \frac{1}{x^2+7} \cdot(2 x+0) \\ & =\frac{-2 x \cdot \sin \left[2 \log \left(x^2+7\right)\right]}{x^2+7} .\end{aligned}$
$e^{3 \sin 2 x-2 \cos 2 x}$
Differentiating w.r.t. x, we get
$\begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}\left[e^{3 \sin ^2 x-2 \cos ^2 x}\right] \\ & =e^{3 \sin ^2 x-2 \cos ^2 x} \cdot \frac{d}{d x}\left(3 \sin ^2 x-2 \cos ^2 x\right) \\ & =e^{3 \sin ^2 x-2 \cos ^2 x} \cdot\left[3 \frac{d}{d x}(\sin x)^2-2 \frac{d}{d x}(\cos x)^2\right] \\ & =e^{3 \sin ^2 x-2 \cos ^2 x} \cdot\left[3 \times 2 \sin x \cdot \frac{d}{d x}(\sin x)-2 \times 2 \cos x \cdot \frac{d}{d x}(\cos x)\right] \\ & =e^{3 \sin ^2 x-2 \cos ^2 x} \cdot[6 \sin x \cos x-4 \cos x(-\sin x)] \\ & =e^{3 \sin ^2 x-2 \cos ^2 x} \cdot(10 \sin x \cos x) \\ & =5\left(2 \sin x \cos ^2 x\right) \cdot e^{3 \sin 2 x-2 \cos ^2 x} \\ & =5 \sin 2 x \cdot e^{3 \sin ^2 x-2 \cos ^2 x} .\end{aligned}$
$\log \left[\cos \left(x^3-5\right)\right]$
Differentiating w.r.t. x, we get
$\frac{d y}{d x}=\frac{d}{d x}\left\{\log \left[\cos \left(x^3-5\right)\right]\right\}$
$\begin{aligned} & =\frac{1}{\cos \left(x^3-5\right)} \cdot \frac{d}{d x}\left[\cos \left(x^3-5\right)\right] \\ & =\frac{1}{\cos \left(x^3-5\right)} \cdot\left[-\sin \left(x^3-5\right)\right] \cdot \frac{d}{d x}\left(x^3-5\right) \\ & =-\tan \left(x^3-5\right) \times\left(3 x^2-0\right) \\ & =-3 x^2 \tan \left(x^3-5\right) .\end{aligned}$
$\operatorname{cosec}(\sqrt{\cos X})$
Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}[\operatorname{cosec}(\sqrt{\cos x})] \\ & =-\operatorname{cosec}(\sqrt{\cos x}) \cdot \cot (\sqrt{\cos x}) \cdot \frac{d}{d x} \sqrt{\cos x} \\ & =-\operatorname{cosec}(\sqrt{\cos x}) \cdot \cot (\sqrt{\cos x}) \cdot \frac{1}{2 \sqrt{\cos x}} \cdot \frac{d}{d x}(\cos x) \\ & =-\operatorname{cosec}(\sqrt{\cos x}) \cdot \cot (\sqrt{\cos x}) \cdot \frac{1}{2 \sqrt{\cos x}} \cdot(-\sin x) \\ & =\frac{\sin x \cdot \operatorname{cosec}(\sqrt{\cos x}) \cdot \cot (\sqrt{\cos x})}{2 \sqrt{\cos x}} .\end{aligned}$
$5^{\sin 3 x+3}$
Differentiating w.r.t. x, we get
$\frac{d y}{d x}=\frac{d}{d x}\left(5^{\sin ^3 x+3}\right)$
$\begin{aligned} & =5^{\sin ^3 x+3} \cdot \log 5 \cdot \frac{d}{d x}\left(\sin ^3 x+3\right) \\ & =5^{\sin ^3 x+3} \cdot \log 5 \cdot\left[3 \sin ^2 x \cdot \frac{d}{d x}(\sin x)+0\right] \\ & =5^{\sin ^3 x+3} \cdot \log 5 \cdot\left[3 \sin ^2 x \cos x\right] \\ & =3 \sin ^2 x \cos x \cdot 5^{\sin ^3 x+3} \cdot \log 5\end{aligned}$
$\cot ^3\left[\log \left(x^3\right)\right]$
Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[\cot \left(\log x^3\right)\right]^3 \\ & =3\left[\cot \left(\log x^3\right)\right]^2 \cdot \frac{d}{d x}\left[\cot \left(\log x^3\right)\right] \\ & =3 \cot ^2\left[\log \left(x^3\right)\right] \cdot\left[-\operatorname{cosec}^2\left(\log x^3\right)\right] \cdot \frac{d}{d x}\left(\log x^3\right) \\ & =-3 \cot ^2\left[\log \left(x^3\right)\right] \cdot \operatorname{cosec} 2\left[\log \left(x^3\right)\right] \cdot 3 \frac{d}{d x}(\log x) \\ & =-3 \cot ^2\left[\log \left(x^3\right)\right] \cdot \operatorname{cosec}^2\left[\log \left(x^3\right)\right] \cdot 3 \times \frac{1}{x} \\ & =\frac{-9 \operatorname{cosec}^2\left[\log \left(x^3\right)\right] \cdot \cot ^2\left[\log \left(x^3\right)\right]}{x} .\end{aligned}$
$\sqrt{\tan \sqrt{x}}$
Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}(\sqrt{\tan \sqrt{x}}) \\ & =\frac{1}{2 \sqrt{\tan \sqrt{x}}} \cdot \frac{d}{d x}(\tan \sqrt{x}) \\ & =\frac{1}{2 \sqrt{\tan \sqrt{x}}} \times \sec ^2 \sqrt{x} \cdot \frac{d}{d x}(\sqrt{x}) \\ & =\frac{1}{2 \sqrt{\tan \sqrt{x}}} \times \sec ^2 \sqrt{x} \times \frac{1}{2 \sqrt{x}} \\ & =\frac{\sec ^2 \sqrt{x}}{4 \sqrt{x} \sqrt{\tan \sqrt{x}}} .\end{aligned}$
$\log \left[\tan \left(\frac{x}{2}\right)\right]$
Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x} \log \left[\tan \left(\frac{x}{2}\right)\right] \\ & =\frac{1}{\tan \left(\frac{x}{2}\right)} \cdot \frac{d}{d x}\left[\tan \left(\frac{x}{2}\right)\right] \\ & =\frac{1}{\tan \left(\frac{x}{2}\right)} \cdot \sec ^2\left(\frac{x}{2}\right) \cdot \frac{d}{d x}\left(\frac{x}{2}\right)\end{aligned}$
$\begin{aligned} & =\frac{\cos \left(\frac{x}{2}\right)}{\sin \left(\frac{x}{2}\right)} \cdot \frac{1}{\cos ^2\left(\frac{x}{2}\right)} \cdot \frac{1}{2} \times 1 \\ & =\frac{1}{2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)} \\ & =\frac{1}{\sin x}=\operatorname{cosec} x .\end{aligned}$
$\sqrt{e^{(3 x+2)}+5}$
Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[e^{(3 x+2)}+5\right]^{\frac{1}{2}} \\ & =\frac{1}{2}\left[e^{(3 x+2)}+5\right]^{-\frac{1}{2}} \cdot \frac{d}{d x}\left[e^{(3 x+2)}+5\right] \\ & =\frac{1}{2 \sqrt{e^{(3 x+2)}+5}} \cdot\left[e^{(3 x+2)} \cdot \frac{d}{d x}(3 x+2)+0\right] \\ & =\frac{1}{2 \sqrt{e^{(3 x+2)}+5}} \cdot\left[e^{(3 x+2)} \cdot(3 \times 1+0)\right] \\ & =\frac{3 e^{(3 x+2)}}{2 \sqrt{e^{(3 x+2)}+5}} \cdot\end{aligned}$
$\cos \left(x^2+a^2\right)$