How do you use the chain rule to differentiate #y=(-5x^3-3)^3#?

Answer 1

#dy/dx=-45x^2(-5x^3-3)^2#

To differentiate using the #color(blue)"chain rule"#

#color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(dy/dx=(dy)/(du)xx(du)/(dx))color(white)(2/2)|)))#

#"let " u=-5x^3-3rArr(du)/(dx)=-15x^2#

#"and " y=u^3rArr(dy)/(du)=3u^2#

#rArrdy/dx=3u^2(-15x^2)#

change u back into terms of x

#rArrdy/dx=-45x^2(-5x^3-3)^2#

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Answer 2

Evelyn Agard

To differentiate ( y=(-5x^3-3)^3 ) using the chain rule:

Identify the outer function, which is ( u^3 ), where ( u = -5x^3 - 3 ).
Differentiate the outer function with respect to ( u ) to get ( \frac{dy}{du} = 3u^2 ).
Identify the inner function, which is ( -5x^3 - 3 ).
Differentiate the inner function with respect to ( x ) to get ( \frac{du}{dx} = -15x^2 ).
Apply the chain rule: Multiply ( \frac{dy}{du} ) and ( \frac{du}{dx} ).
Substitute ( u = -5x^3 - 3 ) back into the result.

The derivative of ( y=(-5x^3-3)^3 ) is ( \frac{dy}{dx} = 3(-5x^3 - 3)^2(-15x^2) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.