(Solved) Find the derivative by using the chain rule

slimes

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3 years ago Edited: 3 years ago, slimes

Find the derivative by using the chain rule - Solution Help

Hi I would like to know if the answers to these questions are correct. Thanks Slight Smile

*For question 8. The 3 is meant to be an exponent, sorry!**

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bio_man

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slimes Author

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3 years ago

Hey thank you so much bio_man Smiling Face with Open Mouth

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! Smiling Face with Glasses

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bio_man

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3 years ago

Quote from: slimes (3 years ago)

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:

\(y'=\frac{3x\left(2-x\right)}{5\cdot \sqrt[5]{\left(2+3x^2-x^3\right)^4}}\)

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also cool new profile!

Thanks!

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slimes Author

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3 years ago

Quote from: bio_man (3 years ago)

Quote from: slimes (3 years ago)

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: \(y'=\frac{3x\left(2-x\right)}{5\cdot \sqrt[5]{\left(2+3x^2-x^3\right)^4}}\)

Thanks will do fix that!

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