How do you find the derivative of the function $y=cosh^-1((sqrt(x))$ ?

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Answer 1 · 2016-12-30T12:57:53

The answer is $=1/(2sqrtxsqrt(x-1))$

We need

$(sqrtx)'=1/(2sqrtx)$

$(coshx)'=sinhx$

$cosh^2x-sinh^2x=1$

Here, we have

$y=cosh^(-1)(sqrtx)$

Therefore,

$coshy=sqrtx$

Taking the derivatives on both sides

$(coshy)'=(sqrtx)'$

$sinhydy/dx=1/(2sqrtx)$

$dy/dx=1/(2sqrtxsinhy)$

$cosh^2y-sinh^2y=1$

$sinh^2y=cos^2y-1$

$sinh^2y=x-1$

$sinhy=sqrt(x-1)$

Therefore,

$dy/dx=1/(2sqrtxsqrt(x-1))$

How do you find the derivative of the function y=cosh^-1((sqrt(x))y=cosh^-1((sqrt(x))?