# How do you use the chain rule to differentiate #y=(4x^3-7)^4(3x+2)^10#?

I got:

If you can't validly modify your answer to match this one, you've mis-factored somewhere.

You would use the product rule on these terms, and the chain rule while using the product rule.

Product Rule:

Chain Rule:

Therefore, what you have (without the chain rule) is:

Now, incorporate the chain rule to finish this step and get:

Now, group terms before factoring:

Factor out common terms:

Factor a little bit more:

And maybe distribute inner terms:

That's probably simplified enough. Either way, it's correct.

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To differentiate ( y = (4x^3 - 7)^4(3x + 2)^{10} ) using the chain rule, follow these steps:

- Identify the functions within the expression: ( u = 4x^3 - 7 ) and ( v = 3x + 2 ).
- Compute the derivatives of ( u ) and ( v ): ( \frac{du}{dx} ) and ( \frac{dv}{dx} ).
- Apply the chain rule: ( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} + \frac{dy}{dv} \cdot \frac{dv}{dx} ).
- Calculate ( \frac{dy}{du} ) and ( \frac{dy}{dv} ).
- Substitute the values obtained in steps 4 and 2 into the chain rule formula to find ( \frac{dy}{dx} ).

Here are the detailed calculations:

[ \frac{du}{dx} = 12x^2 ] [ \frac{dv}{dx} = 3 ] [ \frac{dy}{du} = 4(4x^3 - 7)^3 \cdot 12x^2 ] [ \frac{dy}{dv} = 10(3x + 2)^9 \cdot 3 ] [ \frac{dy}{dx} = 4(4x^3 - 7)^3 \cdot 12x^2 + 10(3x + 2)^9 \cdot 3 ]

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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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