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

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

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Answer 1 · 2017-05-26T01:24:10

Using the product rule $(1 \times d 2 + 2 \times d 1)$ $(1 xx d2 + 2 xx d1)$ , you would set 1 as $x^{3}$ $x^3$ and 2 as ${(2 x - 5)}^{4}$ $(2x-5)^4$ .

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

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

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

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

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How do you differentiate y=x^3(2x-5)^4y=x3(2x−5)4y=x^3(2x-5)^4?

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