What is the derivative of this function $y=csc^-1(4x^2)$ ?

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Answer 1 · 2017-12-27T02:35:28

$dy/dx=-4/(xsqrt(16x^4-1)$

We first need to work the derivative of $csc^-1(x)$
We can start with this:

$y = csc^-1(x) -> csc(y)=x$

Now differentiate both sides implicitly with respect to $x$ to get:

$-dy/dxcsc(y)cot(y)=1$

$-> dy/dx = -1/(csc(y)cot(y))$

Now, using:

$sin^2(y)+cos^2(y)=1$

Divide this identity through by $sin^2(y)$ to get:

$1+cot^2(y)=csc^2(y) -> cot(y) = sqrt(csc^2(y)-1)$

We can now plug this into our equation for $dy/dx$ to get:

$dy/dx = -1/(csc(y)sqrt(csc^2(y)-1)$

Using $csc(y)=x$ we may now rewrite $dy/dx$ in terms of $x$ :

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

So now that we have the derivative of $csc^(-1)x$ we can now apply the chain rule to obtain the derivative of $y=csc^-1(4x^2)$

$-> dy/dx= -1/((4x^2)sqrt((4x^2)^2-1)).d/dx(4x^2)$

$=-(8x)/((4x^2)sqrt((4x^2)^2-1))$

Now simplify:

$=-4/(xsqrt(16x^4-1)$

What is the derivative of this function y=csc^-1(4x^2)y=csc^-1(4x^2)?