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Title: Find the derivative by using the chain rule - Solution Help Post by: slimes on Feb 23, 2021 Hi I would like to know if the answers to these questions are correct. Thanks :)
*For question 8. The 3 is meant to be an exponent, sorry!** Title: Re: Find the derivative by using the chain rule - Solution Help Post by: bio_man on Feb 23, 2021 Content hidden
Title: Re: Find the derivative by using the chain rule - Solution Help Post by: slimes on Feb 24, 2021 Hey thank you so much bio_man :D so my answer would be considered incorrect if I didn't write in factored form?
For 10) I meant to multiply to the -3x^2, that was a mistake! oops also cool new profile! 8-) Title: Re: Find the derivative by using the chain rule - Solution Help Post by: bio_man on Feb 24, 2021 Hey thank you so much bio_man so my answer would be considered incorrect if I didn't write in factored form? Not incorrect, but I'd ask your teacher to make sure you obey his/her style. Teachers can be really subjective sometimes... Actually, I'm not a big fan of my #10 -- it's not wrong, but can be improved. Here's how... \(y'=\frac{\frac{1}{5}\left(6x-3x^2\right)}{\sqrt[5]{\left(2+3x^2-x^3\right)^4}}\) I'd take it a step further by common factoring out the numerator: \(y'=\frac{\frac{1}{5}\left(3x\right)\left(2-x\right)}{\sqrt[5]{\left(2+3x^2-x^3\right)^4}}\) Simplify: \(y'=\frac{\frac{3x}{5}\left(2-x\right)}{\sqrt[5]{\left(2+3x^2-x^3\right)^4}}\) Simplify some more: \(y'=\frac{3x\left(2-x\right)}{5\cdot \sqrt[5]{\left(2+3x^2-x^3\right)^4}}\) Quote also cool new profile! Thanks! ;D Title: Re: Find the derivative by using the chain rule - Solution Help Post by: slimes on Feb 24, 2021 Hey thank you so much bio_man so my answer would be considered incorrect if I didn't write in factored form? Not incorrect, but I'd ask your teacher to make sure you obey his/her style. Teachers can be really subjective sometimes... Actually, I'm not a big fan of my #10 -- it's not wrong, but can be improved. Here's how... \(y'=\frac{\frac{1}{5}\left(6x-3x^2\right)}{\sqrt[5]{\left(2+3x^2-x^3\right)^4}}\) I'd take it a step further by common factoring out the numerator: \(y'=\frac{\frac{1}{5}\left(3x\right)\left(2-x\right)}{\sqrt[5]{\left(2+3x^2-x^3\right)^4}}\) Simplify: \(y'=\frac{\frac{3x}{5}\left(2-x\right)}{\sqrt[5]{\left(2+3x^2-x^3\right)^4}}\) Simplify some more: \(y'=\frac{3x\left(2-x\right)}{5\cdot \sqrt[5]{\left(2+3x^2-x^3\right)^4}}\) |