Q1: Differentiate the following w.r.t. x
\(\frac{e^{x}}{\sin x}\)
Sol:
Let y=\(\frac{e^{x}}{\sin x}\)
Using quotient rule, we have
\(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\sin x.\frac{\mathrm{d} }{\mathrm{d} x}(e^{x})-e^{x}.\frac{\mathrm{d} }{\mathrm{d} x}(\sin x)}{\sin^{2} x}\) \(=\frac{\sin x.(e^{x})-e^{x}.(\cos x)}{\sin^{2} x}\) \(=\frac{e^{x}(\sin x-\cos x)}{\sin^{2} x},x\neq n\pi ,n\in \mathbb{Z}\)
Q2: Differentiate the following w.r.t. x
\(e^{\sin ^{-1}x}\)
Sol:
Let y=\(y=e^{\sin ^{-1}x}\)
Using chain rule, we have
\(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}(e^{\sin ^{-1}x})\) \(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=e^{\sin^{-1}x}.\frac{\mathrm{d} }{\mathrm{d} x}(\sin^{-1}x)\) \(=e^{\sin^{-1}x}.\frac{1}{\sqrt{1-x^{2}}}\) \(=\frac{e^{\sin^{-1}x}}{\sqrt{1-x^{2}}}\) \(therefore,\; \frac{\mathrm{d} y}{\mathrm{d}x}=\frac{e^{\sin^{-1}x}}{\sqrt{1-x^{2}}},x\in (-1,1)\)
Q3: Differentiate the following w.r.t. x
\(e^{x^{3}}\)
Sol:
Let y=\(e^{x^{3}}\)
Using chain rule, we have
\(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}(e^{x^{3}})=e^{x^{3}}.3x^{2}=3x^{2}.e^{x^{3}}\)
Q4: Differentiate the following w.r.t. x
\(\sin (\tan^{-1} e^{-x})\)
Sol:
Let y=\(\sin (\tan^{-1} e^{-x})\)
Using chain rule, we have
\(\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}[\sin (\tan^{-1} e^{-x})]\) \(=\cos (\tan^{-1} e^{-x}).\frac{\mathrm{d} }{\mathrm{d} x}(\tan^{-1} e^{-x})\) \(=\cos (\tan^{-1} e^{-x}).\frac{1}{1+(e^{-x})^{2}}(\tan^{-1} e^{-x}).\frac{\mathrm{d} }{\mathrm{d} x}(e^-{x})\) \(=\frac{\cos (\tan^{-1} e^{-x})}{1+e^{-2x}}.e^{-x}.\frac{\mathrm{d} }{\mathrm{d} x}(-x)\) \(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=e^{\sin^{-1}x}.\frac{\mathrm{d} }{\mathrm{d} x}(\sin^{-1}x)\) \(=e^{\sin^{-1}x}.\frac{1}{\sqrt{1-x^{2}}}\) \(=\frac{e^{\sin^{-1}x}}{\sqrt{1-x^{2}}}\) \(therefore,\; \frac{\mathrm{d} y}{\mathrm{d}x}=\frac{e^{\sin^{-1}x}}{\sqrt{1-x^{2}}},x\in (-1,1)\)
