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  • Differentiate the following with respect to  <img alt="\tan ^{-1}\left(\frac{2^{x+1} \cdot 3^x}{1-(36)^x}\right

Differentiate the following with respect to x

\tan ^{-1}\left(\frac{2^{x+1} \cdot 3^x}{1-(36)^x}\right)

Option: 1

\frac{2^{x+1} \cdot 6^x \log 6}{1+(36)^x}


Option: 2

\frac{2^{x+1} \cdot 3^x \log 6}{3+(36)^x}


Option: 3

\frac{2^{x+1} \cdot 3^x \log 6}{6+(36)^x}


Option: 4

\frac{2^{x+1} \cdot 3^x \log 6}{1+(36)^x}


Answers (1)

best_answer

Let    y=\tan ^{-1}\left(\frac{2^{x+1} \cdot 3^x}{1-(36)^x}\right)

Simplify the given term,

\begin{aligned} & =\tan ^{-1}\left(\frac{2.2^x \cdot 3^x}{1-(6)^{2 x}}\right) \\ & =\tan ^{-1}\left(\frac{2 \cdot 6^x}{1-(6)^{2 x}}\right) \end{aligned}

Put 6^x=\tan \theta

So that,

\theta=\tan ^{-1}\left(6^x\right)

Now, put the value of  6^x=\tan \theta  

 in the obtained simplified term,

\begin{aligned} & \tan ^{-1}\left(\frac{2.6^x}{1-(6)^{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(6^x\right)

2 \theta=2 \tan ^{-1}\left(6^x\right)

and take derivative,

\begin{aligned} & \frac{d y}{d x}=2 \cdot \frac{1}{1+\left(6^x\right)^2} \cdot \frac{d}{d x}\left(6^x\right) \\ & =\frac{2 \cdot 6^x \log 6}{1+6^{2 x}} \\ & =\frac{2^{x+1} \cdot 3^x \log 6}{1+(36)^x} \end{aligned}

 

 

Posted by

vishal kumar

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