What is the derivative of this function $y=sec^-1(4x)$ ?

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Answer 1 · 2016-10-06T23:50:29

$dy/dx=1/(absxsqrt(16x^2-1))$

You may already know that $d/dxsec^-1(x)=1/(absxsqrt(x^2-1))$ . Then, according to the chain rule, we see that we will be following the rule:

$d/dxsec^-1(f(x))=1/(abs(f(x))sqrt((f(x))^2-1))*f'(x)$

So, for $sec^-1(4x)$ , we see that:

$dy/dx=1/(abs(4x)sqrt((4x)^2-1))*d/dx4x$

$dy/dx=4/(abs(4x)sqrt(16x^2-1))$

Note that $abs(4x)=4absx$ :

$dy/dx=1/(absxsqrt(16x^2-1))$

What is the derivative of this function y=sec^-1(4x)y=sec^-1(4x)?