# How do you use the chain rule to differentiate #y=(2x^3+7)^6(2x+1)^8#?

The first step here is to recognize that we are first going to use the product rule followed by the chain rule when we take the derivative:

Following the product rule, lets do the first half:

Now lets do the other half of product rule:

Now just put them to together:

You can simplify it:

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To differentiate ( y = (2x^3 + 7)^6(2x + 1)^8 ) using the chain rule, you first find the derivative of the outer function multiplied by the derivative of the inner function, then multiply them together.

Let ( u = 2x^3 + 7 ) and ( v = 2x + 1 ).

( \frac{dy}{dx} = \frac{d}{dx}[(2x^3 + 7)^6(2x + 1)^8] )

Apply the chain rule:

( \frac{dy}{dx} = \frac{d}{du}(u^6) \cdot \frac{du}{dx} \cdot v^8 + u^6 \cdot \frac{d}{dv}(v^8) \cdot \frac{dv}{dx} )

Differentiate each part:

( \frac{dy}{du} = 6u^5 )

( \frac{du}{dx} = 6x^2 )

( \frac{dy}{dv} = 8v^7 )

( \frac{dv}{dx} = 2 )

Plug the values back in:

( \frac{dy}{dx} = (6(2x^3 + 7)^5) \cdot (6x^2) \cdot (2x + 1)^8 + (2x^3 + 7)^6 \cdot (8(2x + 1)^7) \cdot 2 )

( \frac{dy}{dx} = 12(2x^3 + 7)^5 \cdot 6x^2 \cdot (2x + 1)^8 + 2(2x^3 + 7)^6 \cdot 8(2x + 1)^7 )

Simplify the expression.

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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.

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.

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.

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