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How do you differentiate #y=x^3(2x-5)^4#?

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May 26, 2017

Using the product rule #(1 xx d2 + 2 xx d1)#, you would set 1 as #x^3# and 2 as #(2x-5)^4#.

The derivative of 1 would be #3x^2# and the derivative of 2 is #4(2x-5)^3 *2#, or #8(2x-5)^3# using the chain rule.

Now plug in the values into the product rule, so that #(x^3 xx 8(2x-5)^3) + ((2x-5)^4) xx 3x^2)#.

This expanded is #112 x^6 - 960 x^5 + 3000 x^4 - 4000 x^3 + 1875 x^2#.

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