Question
If $y =\cos ^{-1}\left[\sin \left(4^{ x }\right)\right]$, find $\frac{ d y}{ d x}$
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$x=a t^2, y=2 a t$
$\frac{3 e^{2 x}+5}{4 e^{2 x}-5}$
$\int \frac{1}{3-2 \cos 2 x} \cdot d x$
$x+\frac{d^2 y}{d x^2}=\sqrt{1+\left(\frac{d^2 y}{d x^2}\right)^2}$
$\int \frac{1}{1+x-x^2} \cdot d x$