Chain Rule

[rocking_monkey]

Hi. So i have read from the forums, but now starting to post.

Doing a chain rule. I understand the concept, but I think its more of the algebra that I am stuck on.

*find the derivative of the function
y=x √(1-x^2) so I have

y' = 2x²(1-x²)^1/2 + (1-x² )^1/2 (1)


somehow the next step in the book is (1-x^1/2) [-x² + (1-x²)]
can someone explain to me how it goes from what I have in blue to the one above?

any guidance will be appreciated. Thanks.

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[DrPhil]

Hi. So i have read from the forums, but now starting to post.

Doing a chain rule. I understand the concept, but I think its more of the algebra that I am stuck on.

*find the derivative of the function
y=x√(1-x^2) so I have

y' = 2x²(1-x²)^1/2 + (1-x² )^1/2 (1)


somehow the next step in the book is (1-x^1/2) [-x² + (1-x²)]
can someone explain to me how it goes from what I have in blue to the one above?

any guidance will be appreciated. Thanks.

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Your derivative is not complete - you have to use power law on the square root and then differentiate the argument of the square root. You have combined a lot of steps..

\(\displaystyle \displaystyle y = x\ \left(1-x^2\right)^{1/2}\)

\(\displaystyle \displaystyle y' = x\ \dfrac 12 \left(1-x^2\right)^{-1/2} (-2x) + \left(1-x^2\right)^{1/2}
= \dfrac{-x^2}{\left(1-x^2\right)^{1/2}} + \left(1-x^2\right)^{1/2} \)

....\(\displaystyle = \dfrac{-x^2 + (1 - x^2)}{\left(1-x^2\right)^{1/2}} =\ \cdot\ \cdot\ \cdot \)

Not the same as what you entered from the book - you need to check MY work, too!

 

[rocking_monkey]

Thanks.

Your derivative is not complete - you have to use power law on the square root and then differentiate the argument of the square root. You have combined a lot of steps..


Thank you for the help. I was jumping ahead..