How do you find the derivative of #arcsin (x/2)#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Monzur R. Jun 30, 2017 #d/dx(arcsin(x/2))=1/sqrt(4-x^2)# Explanation: The derivative of #arcsinu#: #d/dx(arcsinu)=(u')/sqrt(1-u^2)# #therefored/dx(arcsin(x/2))=((x/2)')/sqrt(1-(x/2)^2)=(1/2)/sqrt(1-x^2/4)# #=1/(2sqrt((4-x^2)/4))=1/sqrt(4-x^2)# 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 1314 views around the world You can reuse this answer Creative Commons License