How do you find the derivative of #f(x) = (2x-3) ^ -2#? Calculus Basic Differentiation Rules Summary of Differentiation Rules 1 Answer Eddie Jun 26, 2016 #= - 1/(2(2x-3)) # Explanation: #int \ (2x-3) ^ -2 \ dx# the usual rule is #int \ x^n \ dx = (x^(n+1))/(n+1)# here we have #int \ (2x-3) ^ -2 \ dx# with sub #z = 2x-3# to make it more transparent, and so that #dz = 2 dx# the integral becomes #1/2 int \ z ^ -2 \ dz# #= 1/2 int \ (z ^( (-2 + 1)))/(-2 + 1) \ dz# #= 1/2 \ (z ^( -1))/(-1) # #= -1/2 1/z # #= - 1/(2(2x-3)) # Answer link Related questions What is a summary of Differentiation Rules? What are the first three derivatives of #(xcos(x)-sin(x))/(x^2)#? How do you find the derivative of #(e^(2x) - e^(-2x))/(e^(2x) + e^(-2x))#? How do I find the derivative of #y= x arctan (2x) - (ln (1+4x^2))/4#? How do you find the derivative of #y = s/3 + 5s#? What is the second derivative of #(f * g)(x)# if f and g are functions such that #f'(x)=g(x)#... How do you calculate the derivative for #g(t)= 7/sqrtt#? Can you use a calculator to differentiate #f(x) = 3x^2 + 12#? What is the derivative of #ln(x)+ 3 ln(x) + 5/7x +(2/x)#? How do you find the formula for the derivative of #1/x#? See all questions in Summary of Differentiation Rules Impact of this question 2802 views around the world You can reuse this answer Creative Commons License