How do you use the chain rule to differentiate #y=1/(2x^3+7)#?

Answer 1

Ezra Adams

the derivative is #dy/dx=(-6x^2)/((2x^2+7)^2)#

Let #u(x)=2x^3+7#

then #y=(u(x))^(-1)#

so the derivative is #dy/dx=-1(u(x))^(-2)*u'(x)#

we use the following #(x^n)'=nx^(n-1)#

and #u'(x)=(2x^3+7)'=6x^2#

So the final result is #(-1/(2x^3+7)^2)*6x^2#

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Answer 2

Emma Aldridge

To differentiate $y = \frac{1}{2x^3 + 7}$ using the chain rule, follow these steps:

Identify the outer function: In this case, it's the reciprocal function $y = \frac{1}{u}$ , where $u = 2x^3 + 7$ .
Find the derivative of the outer function: The derivative of $\frac{1}{u}$ with respect to $u$ is $-\frac{1}{u^2}$ .
Identify the inner function: The inner function here is $u = 2x^3 + 7$ .
Find the derivative of the inner function: The derivative of $2x^3 + 7$ with respect to $x$ is $6x^2$ .
Apply the chain rule: Multiply the derivative of the outer function by the derivative of the inner function.
Substitute the derivatives into the chain rule formula: $\frac{dy}{dx} = -\frac{1}{(2x^3 + 7)^2} \times (6x^2)$
Simplify the expression if needed.

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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.