Question

​Differentiate

(i) $\left(x^x\right)\sqrt{x}$

(ii) $x^{\left(\sin x-\cos x\right)}+\frac{x^2-1}{x^2+1}$

(iii) $x^{x\cos x+}\frac{x^2+1}{x^2-1}$

(iv) ${\left(x\cos x\right)}^x+{\left(x\sin x\right)}^{1/x}$

(v) ${\left(x+\frac{1}{x}\right)}^x+x^{\left(1+\frac{1}{x}\right)}$

(vi) $e^{\sin x}+{\left(\tan x\right)}^x$

(vii) ${\left(\cos x\right)}^x+{\left(\sin x\right)}^{1/x}$

(viii) $x^{x^2-3}+{{\left(x-3\right)}^x}^2$

Answer 1

(i)
$$\begin{gathered} \mathrm{Let}y=x^x\sqrt{x}...\left(i\right) \\ \mathrm{Taking}\log \mathrm{on}\mathrm{both}\mathrm{sides}, \\ \log y=\log \left(x^x\sqrt{x}\right) \\ \Rightarrow \log y=\log x^x+\log x^{\frac{1}{2}} \\ \Rightarrow \log y=x\log x+\frac{1}{2}\log x \end{gathered}$$
Differentiating with respect to x,
$$\begin{gathered} \frac{1}{y}\frac{dy}{dx}=x\frac{d}{dx}\left(\log x\right)+\log x\frac{d}{dx}\left(x\right)+\frac{1}{2}\frac{d}{dx}\left(\log x\right) \\ \Rightarrow \frac{1}{y}\frac{dy}{dx}=x\left(\frac{1}{x}\right)+\log x\left(1\right)+\frac{1}{2}\left(\frac{1}{x}\right) \\ \Rightarrow \frac{1}{y}\frac{dy}{dx}=1+\log x+\frac{1}{2x} \\ \Rightarrow \frac{dy}{dx}=y\left[1+\log x+\frac{1}{2x}\right] \\ \Rightarrow \frac{dy}{dx}=x^x\sqrt{x}\left[1+\log x+\frac{1}{2x}\right]\left[\mathrm{using}\mathrm{equation}\left(\mathrm{i}\right)\right] \\ \Rightarrow \frac{dy}{dx}=x^{x+\frac{1}{2}}\left[\left(\frac{2x+1}{2x}\right)+\log x\right] \end{gathered}$$


$$\begin{gathered} \left(ii\right)\mathrm{Let}y=x^{\left(\sin x-\cos x\right)}+\left(\frac{x^2-1}{x^2+1}\right) \\ \Rightarrow y=e^{\log x^{\left(\sin x-\cos x\right)}}+\left(\frac{x^2-1}{x^2+1}\right) \\ \Rightarrow y=e^{\left(\sin x-\cos x\right)\log x}+\left(\frac{x^2-1}{x^2+1}\right) \end{gathered}$$
Differentiate it with respect to x using chain rule,
$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}\left[e^{\left(\sin x-\cos x\right)\log x}\right]+\frac{d}{dx}\left[\frac{x^2-1}{x^2+1}\right] \\ =e^{\left(\sin x-\cos x\right)\log x}\frac{d}{dx}\left\{\left(\sin x-\cos x\right)\log x\right\}+\left[\frac{\left(x^2+1\right){\displaystyle \frac{d}{dx}}\left(x^2-1\right)-\left(x^2-1\right){\displaystyle \frac{d}{dx}}\left(x^2+1\right)}{{\left(x^2+1\right)}^2}\right] \\ =e^{\log x^{\left(\sin x-\cos x\right)}}\left[\left(\sin x-\cos x\right)\frac{d}{dx}\left(\log x\right)+\left(\log x\right)\frac{d}{dx}\left(\sin x-\cos x\right)\right]+\left[\frac{\left(x^2+1\right)\left(2x\right)-\left(x^2-1\right)\left(2x\right)}{{\left(x^2+1\right)}^2}\right] \\ =x^{\left(\sin x-\cos x\right)}\left[\left(\sin x-\cos x\right)\left(\frac{1}{x}\right)+\log x\left(\sin x+\cos x\right)\right]+\left[\frac{2x^3+2x-2x^3+2x}{{\left(x^2+1\right)}^2}\right] \\ =x^{\left(\sin x-\cos x\right)}\left[\frac{\left(\sin x-\cos x\right)}{x}+\left(\sin x+\cos x\right)\log x\right]+\frac{4x}{{\left(x^2+1\right)}^2} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{Let}y=x^{x\cos x}+\frac{x^2+1}{x^2-1} \\ \mathrm{Also},\mathrm{Let}u=x^{x\cos x}\mathrm{and}v=\frac{x^2+1}{x^2-1} \\ \therefore y=u+v \\ \Rightarrow \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}...\left(i\right) \\ \mathrm{Now},u=x^{x\cos x} \\ \Rightarrow \log u=\log \left(x^{x\cos x}\right) \\ \Rightarrow \log u=x\cos x\log x \end{gathered}$$
Differentiating both sides with respect to x,
$$\begin{gathered} \frac{1}{u}\frac{du}{dx}=\cos x\log x\frac{d}{dx}\left(x\right)+x\log x\frac{d}{dx}\left(\cos x\right)+x\cos x\frac{d}{dx}\left(\log x\right) \\ \Rightarrow \frac{du}{dx}=u\left[\cos x\log x+x\left(-\sin x\right)\log x+x\cos x\left(\frac{1}{x}\right)\right] \\ \Rightarrow \frac{du}{dx}=x^{x\cos x}\left(\cos x\log x-x\sin x\log x+\cos x\right) \\ \Rightarrow \frac{du}{dx}=x^{x\cos x}\left[\cos x\left(1+\log x\right)-x\sin x\log x\right]...\left(2\right) \\ \mathrm{Again},v=\frac{x^2+1}{x^2-1} \\ \Rightarrow \log v=\log \left(x^2+1\right)-\log \left(x^2-1\right) \end{gathered}$$
Differentiating both sides with respect to x,
$$\begin{gathered} \frac{1}{v}\frac{dv}{dx}=\frac{2x}{x^2+1}-\frac{2x}{x^2-1} \\ \Rightarrow \frac{dv}{dx}=v\left[\frac{2x\left(x^2-1\right)-2x\left(x^2+1\right)}{\left(x^2+1\right)\left(x^2-1\right)}\right] \\ \Rightarrow \frac{dv}{dx}=\frac{x^2+1}{x^2-1}\left[\frac{-4x}{\left(x^2+1\right)\left(x^2-1\right)}\right] \\ \Rightarrow \frac{dv}{dx}=\frac{-4x}{{\left(x^2-1\right)}^2}...\left(3\right) \\ \mathrm{From}\left(\mathrm{i}\right),\left(\mathrm{ii}\right)\mathrm{and}\left(\mathrm{iii}\right),\mathrm{we}\mathrm{obtain} \\ \frac{dy}{dx}=x^{x\cos x}\left[\cos x\left(1+\log x\right)-x\sin x\log x\right]-\frac{4x}{{\left(x^2-1\right)}^2} \end{gathered}$$

$$\begin{gathered} \left(iv\right)\mathrm{Let}y={\left(x\cos x\right)}^x+{\left(x\sin x\right)}^{\frac{1}{x}} \\ \mathrm{Also},\mathrm{Let}u={\left(x\cos x\right)}^x\mathrm{and}v={\left(x\sin x\right)}^{\frac{1}{x}} \\ \therefore y=u+v \\ \Rightarrow \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}...\left(i\right) \\ \mathrm{Now},u={\left(x\cos x\right)}^x \\ \Rightarrow \log u=\log {\left(x\cos x\right)}^x \\ \Rightarrow \log u=x\log \left(x\cos x\right) \\ \Rightarrow \log u=x\left[\log x+\log \cos x\right] \\ \Rightarrow \log u=x\log x+x\log \cos x \end{gathered}$$
Differentiate both sides with respect to x,
$$\begin{gathered} \frac{1}{u}\frac{du}{dx}=\frac{d}{dx}\left(x\log x\right)+\frac{d}{dx}\left(x\log \cos x\right) \\ \Rightarrow \frac{du}{dx}=u\left[\left\{\log x\frac{d}{dx}\left(x\right)+x\frac{d}{dx}\left(\log x\right)\right\}+\left\{\log \cos x\frac{d}{dx}\left(x\right)+x\frac{d}{dx}\left(\log \cos x\right)\right\}\right] \\ \Rightarrow \frac{du}{dx}={\left(x\cos x\right)}^x\left[\left(\log x\left(1\right)+x\left(\frac{1}{x}\right)\right)+\left\{\log \cos x\left(1\right)+x\frac{1}{\cos x}\frac{d}{dx}\left(\cos x\right)\right\}\right] \\ \Rightarrow \frac{du}{dx}={\left(x\cos x\right)}^x\left[\left(\log x+1\right)+\left\{\log \cos x+\frac{x}{\cos x}\left(-\sin x\right)\right\}\right] \\ \Rightarrow \frac{du}{dx}={\left(x\cos x\right)}^x\left[\left(1+\log x\right)+\left(\log \cos x-x\tan x\right)\right] \\ \Rightarrow \frac{du}{dx}={\left(x\cos x\right)}^x\left[1-x\tan x+\left(\log x+\log \cos x\right)\right] \\ \Rightarrow \frac{du}{dx}={\left(x\cos x\right)}^x\left[1-x\tan x+\log \left(x\cos x\right)\right]...\left(ii\right) \\ \mathrm{Again},v={\left(x\sin x\right)}^{\frac{1}{x}} \\ \Rightarrow \log v=\log {\left(x\sin x\right)}^{\frac{1}{x}} \\ \Rightarrow \log v=\frac{1}{x}\log \left(x\sin x\right) \\ \Rightarrow \log v=\frac{1}{x}\left(\log x+\log \sin x\right) \\ \Rightarrow \log v=\frac{1}{x}\log x+\frac{1}{x}\log \sin x \end{gathered}$$
Differentiating both sides with respect to x,
$$\begin{gathered} \frac{1}{v}\frac{dv}{dx}=\frac{d}{dx}\left(\frac{1}{x}\log x\right)+\frac{d}{dx}\left[\frac{1}{x}\log \left(\sin x\right)\right] \\ \Rightarrow \frac{1}{v}\frac{dv}{dx}=\left[\log x\frac{d}{dx}\left(\frac{1}{x}\right)+\frac{1}{x}\frac{d}{dx}\left(\log x\right)\right]+\left[\log \left(\sin x\right)\frac{d}{dx}\left(\frac{1}{x}\right)+\frac{1}{x}\frac{d}{dx}\left\{\log \left(\sin x\right)\right\}\right] \\ \Rightarrow \frac{1}{v}\frac{dv}{dx}=\left[\log x\left(-\frac{1}{x^2}\right)+\left(\frac{1}{x}\right)\left(\frac{1}{x}\right)\right]+\left[\log \left(\sin x\right)\left(-\frac{1}{x^2}\right)+\frac{1}{x}\left(\frac{1}{\sin x}\right)\frac{d}{dx}\left(\sin x\right)\right] \\ \Rightarrow \frac{1}{v}\frac{dv}{dx}=\frac{1}{x^2}\left(1-\log x\right)+\left[-\frac{\log \left(\sin x\right)}{x^2}+\frac{1}{x\sin x}\left(\cos x\right)\right] \\ \Rightarrow \frac{dv}{dx}={\left(x\sin x\right)}^{\frac{1}{x}}\left[\frac{1-\log x}{x^2}+\frac{-\log \left(\sin x\right)+x\cot x}{x^2}\right] \\ \Rightarrow \frac{dv}{dx}={\left(x\sin x\right)}^{\frac{1}{x}}\left[\frac{1-\log x-\log \left(\sin x\right)+x\cot x}{x^2}\right] \\ \Rightarrow \frac{dv}{dx}={\left(x\sin x\right)}^{\frac{1}{x}}\left[\frac{1-\log \left(x\sin x\right)+x\cot x}{x^2}\right]...\left(iii\right) \\ \mathrm{From}\left(\mathrm{i}\right),\left(\mathrm{ii}\right)\mathrm{and}\left(\mathrm{iii}\right),\mathrm{we}\mathrm{obtain} \\ \frac{dy}{dx}={\left(x\cos x\right)}^x\left[1-x\tan x+\log \left(x\cos x\right)\right]+{\left(x\sin x\right)}^{\frac{1}{x}}\left[\frac{x\cot x+1-\log \left(x\sin x\right)}{x^2}\right] \end{gathered}$$

$$\begin{gathered} \left(\mathrm{v}\right)\mathrm{Let}y={\left(x+\frac{1}{x}\right)}^x+x^{\left(1+\frac{1}{x}\right)} \\ \mathrm{Also},\mathrm{Let}u={\left(x+\frac{1}{x}\right)}^x\mathrm{and}v=x^{\left(1+\frac{1}{x}\right)} \\ \therefore y=u+v \\ \Rightarrow \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}...\left(i\right) \\ \mathrm{Then},u={\left(x+\frac{1}{x}\right)}^x \\ \Rightarrow \log u=\log {\left(x+\frac{1}{x}\right)}^x \\ \Rightarrow \log u=x\log \left(x+\frac{1}{x}\right) \end{gathered}$$
Differentiate both sides with respect to x,
$$\begin{gathered} \frac{1}{u}\frac{du}{dx}=\log \left(x+\frac{1}{x}\right)\frac{d}{dx}\left(x\right)+x\frac{d}{dx}\left[\log \left(x+\frac{1}{x}\right)\right] \\ \Rightarrow \frac{1}{u}\frac{du}{dx}=\log \left(x+\frac{1}{x}\right)+x\frac{1}{\left(x+{\displaystyle \frac{1}{x}}\right)}\frac{d}{dx}\left(x+\frac{1}{x}\right) \\ \Rightarrow \frac{du}{dx}=u\left[\log \left(x+\frac{1}{x}\right)+\frac{x}{\left(x+{\displaystyle \frac{1}{x}}\right)}\times \left(1-\frac{1}{x^2}\right)\right] \\ \Rightarrow \frac{du}{dx}={\left(x+\frac{1}{x}\right)}^x\left[\log \left(x+\frac{1}{x}\right)+\frac{\left(x-{\displaystyle \frac{1}{x}}\right)}{\left(x+{\displaystyle \frac{1}{x}}\right)}\right] \\ \Rightarrow \frac{du}{dx}={\left(x+\frac{1}{x}\right)}^x\left[\log \left(x+\frac{1}{x}\right)+\frac{x^2-1}{x^2+1}\right] \\ \Rightarrow \frac{du}{dx}={\left(x+\frac{1}{x}\right)}^x\left[\frac{x^2-1}{x^2+1}+\log \left(x+\frac{1}{x}\right)\right] \\ \mathrm{Again},v=x^{\left(1+\frac{1}{x}\right)} \\ \Rightarrow \log v=\log \left[x^{\left(1+\frac{1}{x}\right)}\right] \\ \Rightarrow \log v=\left(1+\frac{1}{x}\right)\log x \end{gathered}$$
Differentiating both sides with respect to x,
$$\begin{gathered} \frac{1}{v}\frac{dv}{dx}=\log x\frac{d}{dx}\left(1+\frac{1}{x}\right)+\left(1+\frac{1}{x}\right)\frac{d}{dx}\log x \\ \Rightarrow \frac{1}{v}\frac{dv}{dx}=\left(-\frac{1}{x^2}\right)\log x+\left(1+\frac{1}{x}\right)\left(\frac{1}{x}\right) \\ \Rightarrow \frac{1}{v}\frac{dv}{dx}=-\frac{\log x}{x^2}+\frac{1}{x}+\frac{1}{x^2} \\ \Rightarrow \frac{dv}{dx}=v\left[\frac{-\log x+x+1}{x^2}\right] \\ \Rightarrow \frac{dv}{dx}=x^{\left(1+\frac{1}{x}\right)}\left(\frac{x+1-\log x}{x^2}\right)...\left(iii\right) \\ \mathrm{From}\left(\mathrm{i}\right),\left(\mathrm{ii}\right)\mathrm{and}\left(\mathrm{iii}\right),\mathrm{we}\mathrm{obtain} \\ \frac{dy}{dx}={\left(x+\frac{1}{x}\right)}^x\left[\frac{x^2-1}{x^2+1}+\log \left(x+\frac{1}{x}\right)\right]+x^{\left(1+\frac{1}{x}\right)}\left(\frac{x+1-logx}{x^2}\right) \end{gathered}$$

$$\begin{gathered} \left(\mathrm{vi}\right)\mathrm{Let}y=e^{\sin x}+{\left(\tan x\right)}^x \\ \Rightarrow y=e^{\sin x}+e^{\log {\left(\tan x\right)}^x} \\ \Rightarrow y=e^{\sin x}+e^{x\log \left(\tan x\right)} \end{gathered}$$
Differentiating with respect to x,
$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}\left(e^{\sin x}\right)+\frac{d}{dx}\left\{e^{x\log \left(\tan x\right)}\right\} \\ =e^{\sin x}\frac{d}{dx}\left(\sin x\right)+e^{x\log \left(\tan x\right)}\frac{d}{dx}\left(x\log \tan x\right) \\ =e^{\sin x}\left(\cos x\right)+e^{\log {\left(\tan x\right)}^x}\left[x\frac{d}{dx}\left(\log \tan x\right)+\log \tan x\frac{d}{dx}\left(x\right)\right] \\ =e^{\sin x}\left(\cos x\right)+{\left(\tan x\right)}^x\left[\frac{x}{\tan x}\left({\sec }^{2}x\right)+\log \tan x\right] \\ =e^{\sin x}\left(\cos x\right)+{\left(\tan x\right)}^x\left[x\sec x\mathrm{cosec}x+\log \tan x\right] \end{gathered}$$

$$\begin{gathered} \left(viii\right)\mathrm{Let}y={\left(\cos x\right)}^x+{\left(\sin x\right)}^{\frac{1}{x}} \\ \Rightarrow y=e^{\log {\left(\cos x\right)}^x}+e^{\log {\left(\sin x\right)}^{\frac{1}{x}}} \\ \Rightarrow y=e^{x\log \left(\cos x\right)}+e^{\frac{1}{x}\log \sin x} \end{gathered}$$
Differentiating with respect to x,
$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}\left(e^{x\log \cos x}\right)+\frac{d}{dx}\left(e^{\frac{1}{x}\log \sin x}\right) \\ =e^{x\log \cos x}\times \frac{d}{dx}\left(x\log \cos x\right)+e^{\frac{1}{x}\log \sin x}\frac{d}{dx}\left(\frac{1}{x}\log \sin x\right) \\ =e^{\log {\left(\cos x\right)}^x}\times \left[x\frac{d}{dx}\left(\log \cos x\right)+\log \cos x\times \frac{d}{dx}\left(x\right)\right]+e^{\log {\left(\sin x\right)}^{\frac{1}{x}}}\times \left[\frac{1}{x}\frac{d}{dx}\left(\log \sin x\right)+\log \sin x\frac{d}{dx}\left(\frac{1}{x}\right)\right] \\ ={\left(\cos x\right)}^x\left[x\left(\frac{1}{\cos x}\right)\frac{d}{dx}\left(\cos x\right)+\log \cos x\left(1\right)\right]+{\left(\sin \right)}^{\frac{1}{x}}\left[\frac{1}{x}\times \frac{1}{\sin x}\times \frac{d}{dx}\left(\sin x\right)+\log \sin x\left(-\frac{1}{x^2}\right)\right] \\ ={\left(\cos x\right)}^x\left[x\left(\frac{1}{\cos x}\right)\left(-\sin x\right)+\log \cos x\right]+{\left(\sin x\right)}^{\frac{1}{x}}\left[\frac{1}{x}\times \frac{1}{\sin x}\left(\cos x\right)-\frac{1}{x^2}\log \sin x\right] \\ ={\left(\cos x\right)}^x\left[\log \cos x-x\tan x\right]+{\left(\sin x\right)}^{\frac{1}{x}}\left[\frac{\cot x}{x}-\frac{1}{x^2}\log \sin x\right] \end{gathered}$$

$$\begin{gathered} \left(\mathrm{viii}\right)\mathrm{Let}y=x^{x^2-3}+{\left(x-3\right)}^{x^2} \\ \mathrm{Also},\mathrm{let}u=x^{x^2-3}\mathrm{and}v={\left(x-3\right)}^{x^2} \\ \therefore y=u+v \end{gathered}$$
Differentiate both sides with respect to x,
$$\begin{gathered} \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}...\left(i\right) \\ \mathrm{Now},u=x^{x^2-3} \\ \therefore \log u=\log \left(x^{x^2-3}\right) \\ \Rightarrow \log u=\left(x^2-3\right)\log x \end{gathered}$$
Differentiating with respect to x,
$$\begin{gathered} \frac{1}{u}\frac{du}{dx}=\log x\frac{d}{dx}\left(x^2-3\right)+\left(x^2-3\right)\frac{d}{dx}\left(\log x\right) \\ \Rightarrow \frac{1}{u}\frac{du}{dx}=\log x\left(2x\right)+\left(x^2-3\right)\left(\frac{1}{x}\right) \\ \Rightarrow \frac{du}{dx}=x^{x^2-3}\left[\frac{x^2-3}{x}+2x\log x\right] \\ \mathrm{Also},v={\left(x-3\right)}^{x^2} \\ \therefore \log v=\log {\left(x-3\right)}^{x^2} \\ \Rightarrow \log v=x^2\log \left(x-3\right) \end{gathered}$$
Differentiating both sides with respect to x,
$$\begin{gathered} \frac{1}{v}\frac{dv}{dx}=\log \left(x-3\right)\frac{d}{dx}\left(x^2\right)+x^2\frac{d}{dx}\left[\log \left(x-3\right)\right] \\ \Rightarrow \frac{1}{v}\frac{dv}{dx}=\log \left(x-3\right)\left(2x\right)+x^2\left(\frac{1}{x-3}\right)\frac{d}{dx}\left(x-3\right) \\ \Rightarrow \frac{dv}{dx}=v\left[2x\log \left(x-3\right)+\frac{x^2}{x-3}\times 1\right] \\ \Rightarrow \frac{dv}{dx}={\left(x-3\right)}^{x^2}\left[\frac{x^2}{x-3}+2x\log \left(x-3\right)\right] \end{gathered}$$
$$\begin{gathered} \mathrm{Substituing}\mathrm{the}\mathrm{expressions}\mathrm{of}\frac{du}{dx}\mathrm{and}\frac{dv}{dx}\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right) \\ \frac{dy}{dx}=x^{x^2-3}\left[\frac{x^2-3}{x}+2x\log x\right]+{\left(x-3\right)}^{x^2}\left[\frac{x^2}{x-3}+2x\log \left(x-3\right)\right] \end{gathered}$$