How do you find the derivative of the function y = arc cos e^(4x)? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 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)) Answer link Related questions What is the derivative of f(x)=sin^-1(x) ? What is the derivative of f(x)=cos^-1(x) ? What is the derivative of f(x)=tan^-1(x) ? What is the derivative of f(x)=sec^-1(x) ? What is the derivative of f(x)=csc^-1(x) ? What is the derivative of f(x)=cot^-1(x) ? What is the derivative of f(x)=(cos^-1(x))/x ? What is the derivative of f(x)=tan^-1(e^x) ? What is the derivative of f(x)=cos^-1(x^3) ? What is the derivative of f(x)=ln(sin^-1(x)) ? See all questions in Differentiating Inverse Trigonometric Functions Impact of this question 3980 views around the world You can reuse this answer Creative Commons License