Mathematics: This one as well. #22 a & b

Nah but thanks for the points...

This one as well. #22 a & b - id223565280, lolomgwtfnvm4

Ответ:

tyree111

08.01.2022

Mathematics

y' = 90x +33y' = -4/(2x +3)^3

Step-by-step explanation:

The chain rule says ...

dy/dx = (dy/du)·(du/dx)

For these problems that means ...

a. dy/dx = (10u +1)·(3) = 10((3x +1) +1)(3) = 3(30x +11)

dy/dx = 90x + 33

___

b. dy/dx = -2u^-3·(2)

dy/dx = -4/(2x +3)^3

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Ответ:

owenbarrows

08.01.2022

Mathematics

22a. y' = 90x + 33

Step-by-step explanation:

y = 5u² + u - 1 u = 3x + 1

First take the derivative of y = 5u² + u - 1 with respect to u $\bigg(\dfrac{dy}{du}\bigg)$

y' = (2)(5u)(u') + (1)(u') - 0

= (10u)u' + u'

Next, take the derivative of u = 3x + 1 with respect to x $\bigg(\dfrac{du}{dx}\bigg)$

u' = 3 + 0

u' = 3

Now, input u = 3x + 1 and u' = 3 into the y' equation:

y' = (10u)u' + u'

= 10(3x + 1)(3) + 3

= 30(3x + 1) + 3

= 90x + 30 + 3

= 90x + 33

***********************************************************************

22b. $\bold{y'=-\dfrac{4}{(2x+3)^{3}}}$

Step-by-step explanation:

$y=\dfrac{1}{u^2}$ u = 2x + 3

$\text{We can rewrite y as }y=u^{-2}\\\text{Take the derivative of }y=u^{-2}\text{ with respect to u}\ \bigg(\dfrac{dy}{du}\bigg)\\\\y'=(-2)(u^{-3})(u')$

Next, take the derivative of u = 2x + 3 with respect to x $\bigg(\dfrac{du}{dx}\bigg)$

u' = 2 + 0

u' = 2

Now, input u = 2x + 3 and u' = 2 into the y' equation

$y'=(-2)(2x+3)^{-3}(2)\\\\y'=-4(2x+3)^{-3}\\\\y'=-\dfrac{4}{(2x+3)^{3}}$

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This one as well. #22 a & b

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