Solution

Let $y=\dfrac{e^{x}}{\sin x}$

By using the quotient rule, we obtain

$ \begin{aligned} \dfrac{d y}{d x} & =\dfrac{\sin x \dfrac{d}{d x}(e^{x})-e^{x} \dfrac{d}{d x}(\sin x)}{\sin ^{2} x} \\ & =\dfrac{\sin x \cdot(e^{x})-e^{x} \cdot(\cos x)}{\sin ^{2} x} \\ & =\dfrac{e^{x}(\sin x-\cos x)}{\sin ^{2} x}, x \neq n \pi, n \in \mathbf{Z} \end{aligned} $

Solution

Let $y=e^{\sin ^{-1} x}$

By using the chain rule, we obtain

$ \begin{aligned} & \dfrac{d y}{d x}=\dfrac{d}{d x}(e^{\sin ^{-1} x}) \\ & \begin{aligned} \Rightarrow \dfrac{d y}{d x} & =e^{\sin ^{-1} x} \cdot \dfrac{d}{d x}(\sin ^{-1} x) \\ & =e^{\sin ^{-1} x} \cdot \dfrac{1}{\sqrt{1-x^{2}}} \\ & =\dfrac{e^{\sin ^{-1} x}}{\sqrt{1-x^{2}}} \end{aligned} \\ & \therefore \dfrac{d y}{d x}=\dfrac{e^{\sin ^{-1} x}}{\sqrt{1-x^{2}}}, x \in(-1,1) \end{aligned} $

Solution

Let $y=e^{x^{3}}$

By using the chain rule, we obtain

$\dfrac{d y}{d x}=\dfrac{d}{d x}(e^{x^{3}})=e^{x^{3}} \cdot \dfrac{d}{d x}(x^{3})=e^{x^{3}} \cdot 3 x^{2}=3 x^{2} e^{x^{3}}$

Solution

Let $y=\sin (\tan ^{-1} e^{-x})$

By using the chain rule, we obtain

$ \begin{aligned} \dfrac{d y}{d x} & =\dfrac{d}{d x}[\sin (\tan ^{-1} e^{-x})] \\ & =\cos (\tan ^{-1} e^{-x}) \cdot \dfrac{d}{d x}(\tan ^{-1} e^{-x}) \\ & =\cos (\tan ^{-1} e^{-x}) \cdot \dfrac{1}{1+(e^{-x})^{2}} \cdot \dfrac{d}{d x}(e^{-x}) \\ & =\dfrac{\cos (\tan ^{-1} e^{-x})}{1+e^{-2 x}} \cdot e^{-x} \cdot \dfrac{d}{d x}(-x) \\ & =\dfrac{e^{-x} \cos (\tan ^{-1} e^{-x})}{1+e^{-2 x}} \times(-1) \\ & =\dfrac{-e^{-x} \cos (\tan ^{-1} e^{-x})}{1+e^{-2 x}} \end{aligned} $

Solution

Let $y=\log (\cos e^{x})$

By using the chain rule, we obtain

$ \begin{aligned} \dfrac{d y}{d x} & =\dfrac{d}{d x}[\log (\cos e^{x})] \\ & =\dfrac{1}{\cos e^{x}} \cdot \dfrac{d}{d x}(\cos e^{x}) \\ & =\dfrac{1}{\cos e^{x}} \cdot(-\sin e^{x}) \cdot \dfrac{d}{d x}(e^{x}) \\ & =\dfrac{-\sin e^{x}}{\cos e^{x}} \cdot e^{x} \\ & =-e^{x} \tan e^{x}, e^{x} \neq(2 n+1) \dfrac{\pi}{2}, n \in \mathbf{N} \end{aligned} $

Solution

$\dfrac{d}{d x}(e^{x}+e^{x^{2}}+\ldots+e^{x^{3}})$

$=\dfrac{d}{d x}(e^{x})+\dfrac{d}{d x}(e^{x^{2}})+\dfrac{d}{d x}(e^{x^{3}})+\dfrac{d}{d x}(e^{x^{4}})+\dfrac{d}{d x}(e^{x^{3}})$

$=e^{x}+[e^{x^{2}} \times \dfrac{d}{d x}(x^{2})]+[e^{x^{3}} \cdot \dfrac{d}{d x}(x^{3})]+[e^{x^{4}} \cdot \dfrac{d}{d x}(x^{4})]+[e^{x^{5}} \cdot \dfrac{d}{d x}(x^{5})]$

$=e^{x}+(e^{x^{2}} \times 2 x)+(e^{x^{3}} \times 3 x^{2})+(e^{x^{4}} \times 4 x^{3})+(e^{x^{3}} \times 5 x^{4})$

$=e^{x}+2 x e^{x^{2}}+3 x^{2} e^{x^{3}}+4 x^{3} e^{x^{4}}+5 x^{4} e^{x^{5}}$

Solution

Let $y=\sqrt{e^{\sqrt{x}}}$

Then, $y^{2}=e^{\sqrt{x}}$

By differentiating this relationship with respect to $x$, we obtain

$ \begin{aligned} & y^{2}=e^{\sqrt{x}} \\ & \Rightarrow 2 y \dfrac{d y}{d x}=e^{\sqrt{x}} \dfrac{d}{d x}(\sqrt{x}) \quad \text{ [By applying the chain rule] } \\ & \Rightarrow 2 y \dfrac{d y}{d x}=e^{\sqrt{x}} \dfrac{1}{2} \cdot \dfrac{1}{\sqrt{x}} \\ & \Rightarrow \dfrac{d y}{d x}=\dfrac{e^{\sqrt{x}}}{4 y \sqrt{x}} \\ & \Rightarrow \dfrac{d y}{d x}=\dfrac{e^{\sqrt{x}}}{4 \sqrt{e^{\sqrt{x}}} \sqrt{x}} \\ & \Rightarrow \dfrac{d y}{d x}=\dfrac{e^{\sqrt{x}}}{4 \sqrt{x e^{\sqrt{x}}}}, x>0 \end{aligned} $

Solution

Let $y=\log (\log x)$

By using the chain rule, we obtain

$ \begin{aligned} \dfrac{d y}{d x} & =\dfrac{d}{d x}[\log (\log x)] \\ & =\dfrac{1}{\log x} \cdot \dfrac{d}{d x}(\log x) \\ & =\dfrac{1}{\log x} \cdot \dfrac{1}{x} \\ = & \dfrac{1}{x \log x}, x>1 \end{aligned} $

Solution

Let $y=\dfrac{\cos x}{\log x}$

By using the quotient rule, we obtain

$ \begin{aligned} \dfrac{d y}{d x} & =\dfrac{\dfrac{d}{d x}(\cos x) \times \log x-\cos x \times \dfrac{d}{d x}(\log x)}{(\log x)^{2}} \\ & =\dfrac{-\sin x \log x-\cos x \times \dfrac{1}{x}}{(\log x)^{2}} \\ & =\dfrac{-[x \log x \cdot \sin x+\cos x]}{x(\log x)^{2}}, x>0 \end{aligned} $

Solution

Let $y=\cos (\log x+e^{x})$

By using the chain rule, we obtain

$ \begin{aligned} \dfrac{d y}{d x} & =-\sin (\log x+e^{x}) \cdot \dfrac{d}{d x}(\log x+e^{x}) \\ & =-\sin (\log x+e^{x}) \cdot[\dfrac{d}{d x}(\log x)+\dfrac{d}{d x}(e^{x})] \\ & =-\sin (\log x+e^{x}) \cdot(\dfrac{1}{x}+e^{x}) \\ & =-(\dfrac{1}{x}+e^{x}) \sin (\log x+e^{x}), x>0 \end{aligned} $