What is the derivative of #int_(cosx)^(7x) \ cost^3 \ dt # wrt #x#?

Question

Emma Aldridge · Accepted Answer

# d/dx int_(cosx)^(7x) \ cost^3 \ dt = 7 cos(7x)^3 + sinx cos(cos^3x) #

If asked to find the derivative of an integral then you should not evaluate the integral but instead use the fundamental theorem of Calculus. The FTOC tells us that: # d/dx \ int_a^x \ f(t) \ dt = f(x) # for any constant #a# (ie the derivative of an integral gives us the original function back). We are asked to find: # g'(x) = d/dx \ int_(cosx)^(7x) \ cost^3 \ dt# Note that neither the upper or lower bounds are in the correct format for the FTOC to be applied, directly. We can manipulate the definite integral as follows: # int_(cosx)^(7x) \ cost^3 \ dt = int_(a)^(7x) \ cost^3 \ dt - int_a^(cosx) \ cost^3 \ dt # And so: # g'(x) = d/dx int_(a)^(7x) \ cost^3 \ dt - d/dx int_a^(cosx) \ cost^3 \ dt # ... [A] We have arbitrary chosen the lower limit as some arbitrary constant #a# (we could use #0# wlog, any number will do!). We can further manipulate these definite integral using a substitution and the chain rule. Let: # u=7x => (du)/dx = 7 # # v=cosx => (dv)/dx = -sinx # Then substituting into [A], and applying the chain rule, we get: # g'(x) = d/dx int_(a)^u \ cost^3 \ dt - d/dx int_a^v \ cost^3 \ dt # # " " = (du)/dx d/(du) int_(a)^u \ cost^3 \ dt - (dv)/dx d/(dv) int_a^v \ cost^3 \ dt # # " " = 7 d/(du)int_(a)^u \ cost^3 \ dt - (-sinx) d/(dv) int_a^v \ cost^3 \ dt # # " " = 7 d/(du)int_(a)^u \ cost^3 \ dt + sinx d/(dv) int_a^v \ cost^3 \ dt # And now the derivative of both the integrals are in the correct form for the FTOC to be applied, giving: # g'(x) = 7 cosu^3 + sinx cosv^3 # And restoring the initial substitution we get: # g'(x) = 7 cos(7x)^3 + sinx cos(cos^3x) #

Charles Anctil · Answer

undefined The derivative of $ \int_{\cos(x)}^{7x} \cos(t)^3 \, dt $ with respect to $ x $ is $ -7\cos(7x)^3 \cdot 7 + 3\cos(x)^2 \sin(x) $.