You use the product rule with and , and then the chain rule to evaluate the derivative of .
- Hollywood
Done most of my practice problems easily, but this one has me stumped..
y = x^7(3x^4 + 12x - 1)^2
Do I use the Product Rule with f(x) and g(x) both being the term inside the parenthesis, but if so what on earth do I do with the x^7 on the outside? Never had a term in front like this. Any help getting this started would surely be appreciated!
Theoretically, I am answering your question in the same way as Hollywood. But my way may be a bit more intuitive and is far less prone to error for beginners.
$y = x^7(3x^4 + 12x - 1)^2.$
I see a product so product rule comes to mind.
$u = x^7\ and\ v = (3x^4 + 12x - 1)^2 \implies y = uv \implies \dfrac{dy}{dx} = \dfrac{du}{dx} * v + u * \dfrac{dv}{dx}.$ Simple.
$\dfrac{du}{dx} = what?$ There is part of what you need to put into the equation in the previous line.
$v = (3x^4 + 12x - 1)^2.$ I see a power so I think power rule.
$w = 3x^4 + 12x - 1 \implies v = w^2 \implies \dfrac{dv}{dw} = 2w = 2(3x^4 + 12x - 1).$
But I need $\dfrac{dv}{dx}$, not $\dfrac{dv}{dw}.$
I see that w is a function of x so I can find $\dfrac{dw}{dx} = what?$
Now the chain rule let's me compute $\dfrac{dv}{dx} = \dfrac{dv}{dw} * \dfrac{dw}{dx} = what?$
Now put that all together. A lot of differential calculus is simply applying one rule after another. Once you get good at it, you can do it without breaking it down into steps. But it sometimes help to see it done in baby steps, particularly for someone like me who sometimes makes careless mistakes in algebra.
If you have not yet learned the chain rule, then
$\begin{align*}x^7(3x^4+12x-1)^2 & = x^7(9x^8+72x^5-6x^4+144x^2-24x+1) \\ & = 9x^{15}+72x^{12}-6x^{11}+144x^9-24x^8+x^7\end{align*}$
Now you can differentiate term-by-term.