Differentiation. Where am I going wrong?

[Empty_Atoms]

Hey everybody, first of thanks so much to those of you contributing your time and knowledge to these questions.

I'm preparing for a very basic calc exam next month, having trouble with a problem for which I think I should be able to use the chain rule:



I believe I can express this function as a composite function, v(x)= x⁸ and u(x)= (3+2⁴) right?

So I just kinda used the power and constant rules from there to say v prime of u(x) is 7x, and u prime of x is 8x³

In so doing I get 8x⁷ times 8x³... 8¹⁰?


I've tried this in the derivative calculator online and get this though, which makes my head spin...

Can anyone tell me where I'm messing up? Even if I'm unclear in my question, feedback is massively appreciated! Thanks guys and gals!

 

[tkhunny]

Well, these words do not fill one with confidence, "...kinda used the..."

[math]\dfrac{d}{dx}\left(3 + 2\cdot x^{4}\right)^{8} = 8\cdot\left(3 + 2\cdot x^{4}\right)^{7}\cdot\dfrac{d}{dx}\left(3 + 2\cdot x^{4}\right) = 8\cdot\left(3 + 2\cdot x^{4}\right)^{7}\cdot\left(8\cdot x^{3}\right)[/math]
You tell me where you wandered off.

 

[JeffM]

Let's abandon function notation, make a u-substitution, use Leibniz notation for derivatives, and remember the chain rule.

[MATH]y = f(x).\\ \therefore y = (3 + 2x^4)^8.\\ \text {Let } u = 3 + 2x^4 \implies \dfrac{du}{dx} = 8x^3,\ y = u^8, \text { and } \dfrac{dy}{du} = 8u^7.\\ \therefore \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dx} = 8u^7 * 8x^3 = 64x^3u^7 = 64x^3(3 + 2x^4)^7 \\ \therefore f'(x) = 64x^3(3 + 2x^4)^7.[/MATH]There are all sorts of techniques to make calculus less terrifying. Learn them.

 

[Empty_Atoms]

Well, these words do not fill one with confidence, "...kinda used the..."

Are you a teacher by any chance, tkhunny? Rest assured; my aim wasn't to fill you with confidence, and I'm glad the absence thereof came across.

Let's abandon function notation, make a u-substitution, use Leibniz notation for derivatives, and remember the chain rule.

[MATH]y = f(x).\\ \therefore y = (3 + 2x^4)^8.\\ \text {Let } u = 3 + 2x^4 \implies \dfrac{du}{dx} = 8x^3,\ y = u^8, \text { and } \dfrac{dy}{du} = 8u^7.\\ \therefore \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dx} = 8u^7 * 8x^3 = 64x^3u^7 = 64x^3(3 + 2x^4)^7 \\ \therefore f'(x) = 64x^3(3 + 2x^4)^7.[/MATH]There are all sorts of techniques to make calculus less terrifying. Learn them.

This was massively helpful, I see now in that U substitution I'm forgetting what my variables are actually representing. Thank you so much for this I think I have the process now I just need to drill it. So so grateful, thank you.

 

[tkhunny]

Rest assured; my aim wasn't to fill you with confidence, and I'm glad the absence thereof came across.

I deliberately used the impersonal pronoun. I'm not the one who needs the confidence. Part of the learning of mathematics is learning how to communicate it. Feel free to state clearly what it is you are doing. Even if you don't implement perfectly, we will at least know what it is you were trying to do.

BTW Good work coming back and discussing. There are may here who wish to and can help if you come back and discuss.