What is y’ ? y= tan(sin x) + 1/3.412 Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Sonnhard · Stefan V. Jul 21, 2018 y'=(1+tan^2(sin(x))*cos(x) Explanation: We are Using the chain rule (f(g(x))'=f'(g(x))*g'(x) and that (tan(x))'=1+tan^2(x) so we get y'=(1+tan^2(sin(x))*cos(x) 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 1730 views around the world You can reuse this answer Creative Commons License