22a. Answer: 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 [tex]\bigg(\dfrac{dy}{du}\bigg)[/tex]
y' = (2)(5u)(u') + (1)(u') - 0
= (10u)u' + u'
Next, take the derivative of u = 3x + 1 with respect to x [tex]\bigg(\dfrac{du}{dx}\bigg)[/tex]
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
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22b. Answer: [tex]\bold{y'=-\dfrac{4}{(2x+3)^{3}}}[/tex]
Step-by-step explanation:
[tex]y=\dfrac{1}{u^2}[/tex] u = 2x + 3
[tex]\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')[/tex]
Next, take the derivative of u = 2x + 3 with respect to x [tex]\bigg(\dfrac{du}{dx}\bigg)[/tex]
u' = 2 + 0
u' = 2
Now, input u = 2x + 3 and u' = 2 into the y' equation
[tex]y'=(-2)(2x+3)^{-3}(2)\\\\y'=-4(2x+3)^{-3}\\\\y'=-\dfrac{4}{(2x+3)^{3}}[/tex]
