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  • Differentiate the following with respect to   <img alt="\tan ^{-1}\left(\frac{2^{x+1} \cdot 4^x}{1-(64)^x}

Differentiate the following with respect to  x

\tan ^{-1}\left(\frac{2^{x+1} \cdot 4^x}{1-(64)^x}\right)

Option: 1

\frac{2^{x+1} \cdot 8^x \log 8}{1+(64)^x}


Option: 2

\frac{2^{x+1} \cdot 4^x \log 8}{8+(64)^x}


Option: 3

\frac{2^{x+1} \cdot 4^x \log 8}{1+(64)^x}


Option: 4

\frac{2^{x+1} \cdot 4^x \log 8}{4+(64)^x}


Answers (1)

let   y=\tan ^{-1}\left(\frac{2^{x+1} \cdot 4^x}{1-(64)^x}\right)

Simplify the given term,

$$ \begin{aligned} & =\tan ^{-1}\left(\frac{2.2^x \cdot 4^x}{1-(8)^{2 x}}\right) \\ & =\tan ^{-1}\left(\frac{2.8^x}{1-(8)^{2 x}}\right) \end{aligned}

put,8^x=\tan \theta 

so that,

\theta=\tan ^{-1}\left(8^x\right)

so now put the value of 8^x=\tan \theta in the obtained simplified term,

\begin{aligned} & \tan ^{-1}\left(\frac{2.8^x}{1-(8)^{2 x}}\right)=\tan ^{-1}\left(\frac{2 \tan \theta}{1-\tan ^2 \theta}\right) \\ & =\tan ^{-1}(\tan 2 \theta) \\ & =2 \theta \end{aligned}

put the value of   \theta from (1),

2 \theta=2 \tan ^{-1}\left(8^x\right)

and take derivatives,

\begin{aligned} & \frac{d y}{d x}=2 \cdot \frac{1}{1+\left(8^x\right)^2} \cdot \frac{d}{d x}\left(8^x\right) \\ & =\frac{2 \cdot 8^x \log 8}{1+8^{2 x}} \\ & =\frac{2^{x+1} \cdot 4^x \log 8}{1+(64)^x} \end{aligned}

Posted by

Ramraj Saini

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