# Differentiate 𝑦 = (2𝑥^3 − 5𝑥 + 1)^20(3𝑥 − 5)^10

• ttpp1124
In summary: Look at the statement of the problem in post #1.sorry for the late reply, but I think I got it..?Just differentiating the equation via a calculator I got:$$y' = 20\left(2x^3-5x+1\right)^{19}\left(6x^2-5\right)\left(3x-5\right)^{10}+30\left(3x-5\right)^9\left(2x^3-5x+1\right)^{20}$$or $$y' = 10(3x-5)^9 (2x^3 - 5x +1)^{ ttpp1124 Homework Statement Differentiate 𝑦 = (2𝑥^3 − 5𝑥 + 1)^20(3𝑥 − 5)^10. Write your answer in simplest factored form. I've solved it..but I feel like it can be factored further? Relevant Equations n/a Last edited by a moderator: I don't see any reason to use "logarithmic integration". Just the "product rule" and "chain rule" are sufficient. $$y= (2𝑥^3 − 5𝑥 + 1)^{20}(3𝑥 − 5)^{10}$$ $$y'= 20(2x^3- 5x+ 1)^{19}(6x^2- 5)(3x- 5)^{10}+ (2x^3- 5x+ 1)^{20}(10(3x- 5)^{10}(3))$$. Both third degree polynomials you end up with have real roots (they are polynomials of odd degree). I don't expect that you can completely factorise these into linear/quadratic factors as the roots will probably be quite ugly. HallsofIvy said: I don't see any reason to use "logarithmic integration". Just the "product rule" and "chain rule" are sufficient. $$y= (2𝑥^3 − 5𝑥 + 1)^{20}(3𝑥 − 5)^{10}$$ $$y'= 20(2x^3- 5x+ 1)^{19}(6x^2- 5)(3x- 5)^{10}+ (2x^3- 5x+ 1)^{20}(10(3x- 5)^{10}(3))$$. wow, this question can really be solved in one step like that? I guess I will leave it as it is then. Math_QED said: Both third degree polynomials you end up with have real roots (they are polynomials of odd degree). I don't expect that you can completely factorise these into linear/quadratic factors as the roots will probably be quite ugly. ttpp1124 said: wow, this question can really be solved in one step like that? Well, the quicker you try to do it, the more likely you are to make mistakes so perhaps you should check my calculations! SammyS But it isn't in best factored form yet. Nothing was said about factoring! HallsofIvy said: Nothing was said about factoring! Look at the statement of the problem in post #1. sorry for the late reply, but I think I got it..? Just differentiating the equation via a calculator I got:$$y' = 20\left(2x^3-5x+1\right)^{19}\left(6x^2-5\right)\left(3x-5\right)^{10}+30\left(3x-5\right)^9\left(2x^3-5x+1\right)^{20}$$or$$y' = 10(3x-5)^9 (2x^3 - 5x +1)^{19} (42x^3 - 60x^2 - 45x + 53)

If you ever just want to double-check your answers, you could always use either symbolab.com or wolframalpha.com. If your university provides students with the pro (i.e. paid) version of WoframAlpha, you could even check your work. Needless to say, it's highly recommended to do your work by hand first before checking the step-by-step solution process which WoframAlpha Pro provides. Note that Symbolab also has the step-by-step solution process too. However, to my knowledge, WolframAlpha is much more versatile and powerful over all. If anything, though, these online calculators are a great tool to double-check your answers.

If you just write the problem in the first place as differentiate y = u20v10 it is more obvious and easy.

## 1. What is the formula for differentiating 𝑦 = (2𝑥^3 − 5𝑥 + 1)^20(3𝑥 − 5)^10?

The formula for differentiating 𝑦 = (2𝑥^3 − 5𝑥 + 1)^20(3𝑥 − 5)^10 is d𝑦/d𝑥 = 20(2𝑥^3 − 5𝑥 + 1)^19(6𝑥^2 − 5)(3𝑥 − 5)^10 + 10(2𝑥^3 − 5𝑥 + 1)^20(3).

## 2. What is the power rule for differentiating this function?

The power rule for differentiating this function is to multiply the exponent by the coefficient, subtract 1 from the exponent, and then multiply by the derivative of the inside function.

## 3. How do you simplify the derivative of 𝑦 = (2𝑥^3 − 5𝑥 + 1)^20(3𝑥 − 5)^10?

To simplify the derivative, you can use the product rule and the chain rule. First, find the derivative of each term separately, then combine them using the product rule.

## 4. What is the general rule for differentiating a polynomial function?

The general rule for differentiating a polynomial function is to multiply each term by its exponent, subtract 1 from the exponent, and then multiply by the derivative of the inside function. This process is repeated for each term in the polynomial.

## 5. How do you find the critical points of this function?

To find the critical points of this function, you can set the derivative equal to 0 and solve for 𝑥. The resulting values of 𝑥 will be the critical points of the function.

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