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  • Differentiate the following with respect to   <img alt="\tan ^{-1}\left(\frac{2^{x+1} \cdot 2^x}{1-(16)^x}

Differentiate the following with respect to  x

\tan ^{-1}\left(\frac{2^{x+1} \cdot 2^x}{1-(16)^x}\right)

Option: 1

\frac{2^{x+1} \cdot 2^x \log 4}{1+(16)^x}


Option: 2

\frac{2^{x+1} \cdot 4^x \log 4}{4+(16)^x}


Option: 3

\frac{2^{x+1} \cdot 4^x \log 4}{1+(16)^x}


Option: 4

\frac{2^{x+1} \cdot 4 \log 4}{4+(16)^x}


Answers (1)

best_answer

let   y=\tan ^{-1}\left(\frac{2^{x+1} \cdot 2^x}{1-(16)^x}\right)

simplify the given term,

\begin{aligned} & =\tan ^{-1}\left(\frac{2.2^x \cdot 2^x}{1-(4)^{2 x}}\right) \\ & =\tan ^{-1}\left(\frac{2.4^x}{1-(4)^{2 x}}\right) \end{aligned}

put  4^x=\tan \theta

so that ,

\theta=\tan ^{-1}\left(4^x\right)

Now put the value of  4^x=\tan \theta in the obtained simplified term,

\begin{aligned} & \tan ^{-1}\left(\frac{2.4^x}{1-(4)^{2 x}}\right)=\tan ^{-1}\left(\frac{2 \tan \theta}{1-\tan ^2 \theta}\right) \\ & =\tan ^{-1}(\tan 2 \theta) \\ & =2 \theta \end{aligned}

and take derivatives,

\begin{aligned} & \frac{d y}{d x}=2 \cdot \frac{1}{1+\left(4^x\right)^2} \cdot \frac{d}{d x}\left(4^x\right) \\ & =\frac{2 \cdot 4^x \log 4}{1+4^{2 x}} \\ & =\frac{2^{x+1} \cdot 2^x \log 4}{1+(16)^x} \end{aligned}

 

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