Question

Find $\frac{dy}{dx}$

(i) $y=x^{\cos x}+{\left(\sin x\right)}^{\tan x}$
(ii) $y=x^x+{\left(\sin x\right)}^x$
(iii) $y={\left(\sin x\right)}^x+{\sin }^{-1}\sqrt{x}$

Answer 1

$$\begin{gathered} \left(i\right)\mathrm{We}\mathrm{have},y=x^{\cos x}+{\left(\sin x\right)}^{\tan x} \\ \Rightarrow y=e^{\log x^{\cos x}+e^{\log {\left(\sin x\right)}^{\tan x}}} \\ \Rightarrow y=e^{\cos x\log x}+e^{\tan x\log \sin x} \end{gathered}$$
Differentiating with respect to x using chain rule,
$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}\left(e^{\cos x\log x}\right)+\frac{d}{dx}\left(e^{\tan x\log \sin x}\right) \\ =e^{\cos x\log x}\frac{d}{dx}\left(\cos x\log x\right)+e^{\tan x\log \sin x}\frac{d}{dx}\left(\tan x\log \sin x\right) \\ =e^{\log x^{\cos x}}\left[\cos x\frac{d}{dx}\left(\log x\right)+\log x\frac{d}{dx}\left(\cos x\right)\right]+e^{\log {\left(\sin x\right)}^{\tan x}}\left[\tan x\frac{d}{dx}\log \sin x+\log \sin x\frac{d}{dx}\left(\tan x\right)\right] \\ =x^{\cos x}\left[\cos x\left(\frac{1}{x}\right)+\log x\left(-\sin x\right)\right]+{\left(\sin x\right)}^{\tan x}\left[\tan x\left(\frac{1}{\sin x}\right)\frac{d}{dx}\left(\sin x\right)+\log \sin x\left({\sec }^{2}x\right)\right] \\ =x^{\cos x}\left[\frac{\cos x}{x}-\sin x\log x\right]+{\left(\sin x\right)}^{\tan x}\left[\tan x\left(\frac{1}{\sin x}\right)\left(\cos x\right)+{\sec }^{2}x\log \sin x\right] \\ =x^{\cos x}\left[\frac{\cos x}{x}-\sin x\log x\right]+{\left(\sin x\right)}^{\tan x}\left[1+{\sec }^{2}x\log \sin x\right] \end{gathered}$$

$$\begin{gathered} \left(ii\right)\mathrm{We}\mathrm{have},y=x^x+{\left(\sin x\right)}^x \\ \Rightarrow y=e^{\log x^x}+e^{\log {\left(\sin x\right)}^x} \\ \Rightarrow y=e^{x\log x}+e^{x\log \sin x} \end{gathered}$$
Differentiating with respect to x using chain rule and product rule,
$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}\left(e^{x\log x}\right)+\frac{d}{dx}\left(e^{x\log \sin x}\right) \\ =e^{x\log x}\frac{d}{dx}\left(x\log x\right)+e^{x\log \sin x}\frac{d}{dx}\left(x\log \sin x\right) \\ =e^{x\log x}\left[x\frac{d}{dx}\left(\log x\right)+\log x\frac{d}{dx}\left(x\right)\right]+e^{\log {\left(\sin x\right)}^x}\left[x\frac{d}{dx}\left(\log \sin x\right)+\log \sin x\frac{d}{dx}\left(x\right)\right] \\ =x^x\left[x\left(\frac{1}{x}\right)+\log x\left(1\right)\right]+{\left(\sin x\right)}^x\left[x\left(\frac{1}{\sin x}\right)\frac{d}{dx}\left(\sin x\right)+\log \sin x\right] \\ =x^x\left[1+\log x\right]+{\left(\sin x\right)}^x\left[x\left(\frac{1}{\sin x}\right)\left(\cos x\right)+\log \sin x\right] \\ =x^x\left[1+\log x\right]+{\left(\sin x\right)}^x\left[x\cot x+\log \sin x\right] \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{We}\mathrm{have},y={\left(\sin x\right)}^x+{\sin }^{-1}\sqrt{x} \\ \Rightarrow y=e^{\log {\left(\sin x\right)}^x}+{\sin }^{-1}\sqrt{x} \\ \Rightarrow y=e^{x\log \sin x}+{\sin }^{-1}\sqrt{x} \end{gathered}$$
Differentiating with respect to x using chain rule and product rule,
$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}\left(e^{x\log \sin x}\right)+\frac{d}{dx}{\sin }^{-1}\sqrt{x} \\ =e^{x\log \sin x}\frac{d}{dx}\left(x\log \sin x\right)+\frac{1}{\sqrt{1-{\left(\sqrt{x}\right)}^2}}\frac{d}{dx}\left(\sqrt{x}\right) \\ =e^{\log {\left(\sin x\right)}^x}\left[x\frac{d}{dx}\left(\log \sin x\right)+\log \sin x\frac{d}{dx}\left(x\right)+\frac{1}{\sqrt{1-x}}\times \frac{1}{2\sqrt{x}}\right] \\ ={\left(\sin x\right)}^x\left[x\frac{1}{\sin x}\frac{d}{dx}\left(\sin x\right)+\log \sin x\left(1\right)\right]+\frac{1}{2\sqrt{x-x^2}} \\ ={\left(\sin x\right)}^x\left[\left(\frac{x}{\sin x}\right)\left(\cos x\right)+\log \sin x\right]+\frac{1}{2\sqrt{x-x^2}} \\ ={\left(\sin x\right)}^x\left[x\cot x+\log \sin x\right]+\frac{1}{2\sqrt{x-x^2}} \end{gathered}$$