How do you differentiate #f(x)= (4x^2+5)(3x-5) # using the product rule?

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How do you differentiate #f(x)= (4x^2+5)(3x-5) # using the product rule?

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Answer 1 · 2016-03-12T20:17:52

#d/(d x) f(x)=36x^2-40x+15#

Explanation:

#f(x)=(4x^2+5)(3x-5)#
#d/(d x) f(x)=(4x^2+5)^' * (3x-5)+(3x-5) ^ ' * (4x^2+5)#
#d/(d x) f(x)=8x(3x-5)+3(4x^2+5)#
#d/(d x) f(x)=24x^2-40x+12x^2+15#
#d/(d x) f(x)=36x^2-40x+15#

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How do you differentiate #f(x)= (4x^2+5)(3x-5) # using the product rule?

1 Answer

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