Question
Differentiate the following functions with respect to x:
$\sin^{-1}\Big(\frac{1}{\sqrt{1+\text{x}^2}}\Big)$
$\sin^{-1}\Big(\frac{1}{\sqrt{1+\text{x}^2}}\Big)$
✓
Answer
Let $\text{f(x)}=\sin^{-1}\Big(\frac{2^{\text{x}+1}}{1+4^\text{x}}\Big)$ To find the domain, we need to find all x such that $-1\leq\frac{2^{\text{x}+1}}{1+4^\text{x}}\leq1$ Since the quantity in the middle is always psitive, we need to find all x such that $\frac{2^{\text{x}+1}}{1+4^\text{x}}\leq1$ i.e. all x such that $2^{\text{x}+1}\leq1+4^\text{x}$ We may req. write as $2\leq\frac{1}{2^\text{x}}+2^\text{x},$ which is true for all x Hence, the function is defined at all real numbers. Putting $2^\text{x}=\tan\theta$
$\text{f(x)}=\sin^{-1}\Big(\frac{2^{\text{x}+1}}{1+4^\text{x}}\Big)=\sin^{-1}\Big(\frac{2^\text{x}.2}{1+(2^\text{x})^2}\Big)$
$=\sin^{-1}\Big[\frac{2\tan\theta}{1+\tan^2\theta}\Big]=\sin^{-1}(\sin2\theta)=2\theta=2\tan^{-1}(2^\text{x})$ Thus, $\text{f(x)}=2\frac{1}{1+(2^\text{x})^2}\frac{\text{d}}{\text{dx}}(2^\text{x})$ $=\frac{2}{1+4^\text{x}}(2^\text{x})\log2=\frac{2^{\text{x}+1}\log2}{1+4^\text{x}}$Need a full question paper?
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