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

1 Answer
Bill K.
Jun 23, 2015

#dy/dx=-(4e^(4x))/sqrt(1-e^(8x))#

Explanation:

The derivative of #arccos(x)# is #d/dx(arccos(x))=-1/sqrt(1-x^2)# and we also know #d/dx(e^(x))=e^(x)#. 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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