What is the derivative of #f(x)= (2x-3)^3 (x+1)^2#?
2 Answers
Explanation:
Note: this is simply an alternative method to Steve M's version which uses the Product and Chain Rules (and was what was probably intended)
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Given:
and the expansion of
and then the derivative can be found by by noting that the derivative of a sum of terms is simply the sum of the derivatives of the individual terms (and applying the exponent rule for derivatives)
This is simply an expansion of Steve M's version (and you should look at it, if only as an application of the Product and Chain Rules)
# f'(x) = 10x(2x-3)^2(x+1) #
Explanation:
We seek the derivative of:
# f(x) = (2x-3)^3(x+1)^2 #
We can apply the product rule so that:
# f'(x) = (2x-3)^3 \ (d/dx(x+1)^2) + (d/dx(2x-3)^3) \ (x+1)^2 #
And by application of the chain rule, we have:
# d/dx(x+1)^2 = 2(x+1) \ (d/dx(x+1)) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2(x+1)#
Similarly:
# d/dx(2x-3)^3 = 3(2x-3)^2 \ d/dx(2x-3) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 3(2x-3)^2 (2) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 6(2x-3)^2 #
Thus we have:
# f'(x) = (2x-3)^3 \ 2(x+1) + 6(2x-3)^2 \ (x+1)^2 #
# \ \ \ \ \ \ \ \ \= (2x-3)^2(x+1){ 2(2x-3) + 6 (x+1) }#
# \ \ \ \ \ \ \ \ \= (2x-3)^2(x+1)(4x-6 + 6x+6) #
# \ \ \ \ \ \ \ \ \= (2x-3)^2(x+1)(10x) #
# \ \ \ \ \ \ \ \ \= 10x(2x-3)^2(x+1) #