How do you find the derivative of the function y = arc cos e^(4x)y=arccose4x?

1 Answer
Bill K.
Jun 23, 2015

dy/dx=-(4e^(4x))/sqrt(1-e^(8x))dydx=−4e4x√1−e8x

Explanation:

The derivative of arccos(x)arccos(x) is d/dx(arccos(x))=-1/sqrt(1-x^2)ddx(arccos(x))=−1√1−x2 and we also know d/dx(e^(x))=e^(x)ddx(ex)=ex. We can combine these facts, as well as the Chain Rule (d/dx(f(g(x)))=f'(g(x))*g'(x)) to say that, for y=arccos(e^(4x)), we get

dy/dx=-1/sqrt(1-(e^(4x))^2)*d/dx(e^(4x))=-(4e^(4x))/sqrt(1-e^(8x))

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