Differentiate the following with respect to x | vedclass

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(A) Let $f(x) = \sin ^{-1}\left(\frac{2^{x+1}}{1+4^{x}}\right)$.We can rewrite the expression inside the sine inverse as:$f(x) = \sin ^{-1}\left(\frac

Vedclass · Accepted Answer

$\frac{2^{x+1} \log 2}{1+4^{x}}$

Vedclass · Answer

(A) Let $f(x) = \sin ^{-1}\left(\frac{2^{x+1}}{1+4^{x}}ight)$.We can rewrite the expression inside the sine inverse as:$f(x) = \sin ^{-1}\left(\frac{2 \cdot 2^{x}}{1+(2^{x})^{2}}ight)$.Let $2^{x} = 	an 	heta$,then $	heta = 	an ^{-1}(2^{x})$.Substituting this into the function,we get:$f(x) = \sin ^{-1}\left(\frac{2 	an 	heta}{1+	an ^{2} 	heta}ight) = \sin ^{-1}(\sin 2	heta) = 2	heta = 2 	an ^{-1}(2^{x})$.Now,differentiating with respect to $x$ using the chain rule:$f'(x) = 2 \cdot \frac{1}{1+(2^{x})^{2}} \cdot \frac{d}{dx}(2^{x})$.Since $\frac{d}{dx}(a^{x}) = a^{x} \log a$,we have $\frac{d}{dx}(2^{x}) = 2^{x} \log 2$.Therefore,$f'(x) = 2 \cdot \frac{1}{1+4^{x}} \cdot 2^{x} \log 2 = \frac{2^{x+1} \log 2}{1+4^{x}}$.

Differentiate the following with respect to $x$: $\sin ^{-1}\left(\frac{2^{x+1}}{1+4^{x}}\right)$

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