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Using chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

I'm given that $y=u^3+1$ and $u=5-x^2$

So far, I have came to $\frac{d}{du}(u^3+1) \cdot \frac{d}{dx}(5-x^2) = (3u^2)(-2x)$

Then: $(3u^2)(-2x) = -2x(3(5-x^2)^2)$

I'm not sure how to solve further, I can get $-6x$, but how to I deal with the squared part?

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    Nope, I think your done (unless more simplification/expansion is wanted) – 2017-01-18
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    I might need more simplification, though if it's too complicated, perhaps it's best to keep as that? – 2017-01-18
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    Yep, probably good enough as is. – 2017-01-18

1 Answers 1

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This is normally fine, however if you really do want to expand it then you can do it this way:

$$-2x(3(5-x^2)^2)=-6x(5-x^2)^2$$

Then, using the fact that $(a-b)^2=a^2-2ab+b^2$, we obtain:

$$-6x(5^2-2\cdot 5\cdot x^2+x^4)=-6x(25-10x^2+x^4)=-150x+60x^3-6x^5$$