Question
Diffrentiate the following w. r. t. x.
$\cot ^{-1}\left(4^x\right.$
$\cot ^{-1}\left(4^x\right.$
Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[\cot ^{-1}\left(4^x\right)\right] \\ & =\frac{-1}{1+\left(4^x\right)^2} \cdot \frac{d}{d x}\left(4^x\right) \\ & =\frac{-1}{1+4^{2 x}} \times 4^x \log 4 \\ & =-\frac{4^x \log 4}{1+4^{2 x}}\end{aligned}$
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$f(x)=x^2-4 x+3, x \in[1,3]$