First & Second Derivative of $y=x(2x+3)^4$?

How do you find the first and second derivative of $y=x(2x+3)^4$ using the chain rule?

I know the answers I am just not sure how to solve this equation.

When I try to solve this using the product chain rule, $(f'(x)\cdot g(x))+(g'(x)\cdot f(x))$, I get a very different answer than the one I am suppose to.

The answer I am supposed to get is $y'=(2x+3)^3(10x+3)$ and $y''=(2x+3)^2(80x+48)$

• Using the Product Rule, $$y' = (1)(2x + 3)^4 + 4 x (2x+3)^3 (2) =(2x+3)^3 ( 2x + 3 + 8x) = (10 x+3) (2 x+3)^3$$ Can you proceed?
– Moo
Sep 19 '17 at 20:24
• I am confused at how you get from $(1)(2x+3)^4+4x(2x+3)^3(2)$ to $(2x+3)^3(2x+3+8x)$? Sep 19 '17 at 20:32
• Factor out $(2x+3)^3$. Clear?
– Moo
Sep 19 '17 at 20:41
• Thank you! That makes more sense. Sep 19 '17 at 20:43

Let $f(x) = x$ and $g(x) = \left( h(x) \right) ^4$ where $h(x) = 2x+3$

Then $f'(x) = 1$ and $h'(x) = 2$ which gives further $$g'(x) = 4 \left( h(x) \right) ^3 \cdot h'(x) =8(2x+3)^3$$ Use the product chain rule: \begin{align} y' & = f'(x) \cdot g(x) + f(x) \cdot g'(x) = 1 \cdot (2x+3)^4 + x \cdot 8(2x+3)^3 \\ & = (2x+3+8x)(2x+3)^3 = (10x+3)(2x+3)^3 \end{align}

I hope you can do the same for $y''$ :)

• Thank you, I think I see it now. Though am i just to always take the derivatives of f and h before I take the derivative of g? Sep 19 '17 at 20:36
• yes. Indeed, I think that to avoid mistake, you should try to divide your problem into smaller parts, ideally may have the polynomial formula $a_{0} + a_{1}x + a_{2}x^{2} + \cdots$. Also, it would be easier to compute the first derivative only, therefore instead of computing second derivative, you should compute the derivative of the first derivative
– JKay
Sep 19 '17 at 20:56

What you have written is simply the product rule:

$$(f(x)\cdot g(x))' = f'(x)\cdot g(x) + f(x) \cdot g'(x).$$

To obtain the rule for this situation, we need to combine the chain rule,

$$\big(f(g(x))\big)' = f'(g(x))\cdot g'(x)$$

with the product rule to obtain:

\begin{align*} (f(g(x)))\cdot h(x))' &= \big(f(g(x))\big)'\cdot h(x) + f(g(x)) \cdot h'(x)\\ &= f'(g(x))\cdot g'(x) \cdot h(x) + f(g(x)) \cdot h'(x) \end{align*}.

So what are $f, g$ and $h$ here? Identifying them correctly and using the rule we just developed should allow you to arrive at the desired result. (You'll need to do a little algebra to simply the resulting expression.)

Repeating this process will allow you to find the correct second derivative as well. Good luck!

First apply the product rule for $$y(x)=\underbrace{x}_{f(x)} \underbrace{(2x+3)^4}_{g(x)}$$ and then the chain rule for $g(x) = u(v(x))$ with $u(x) = x^4$ and $v(x) = 2x + 3$.

well, what prevents you from using

$$(u\cdot v)' = u'v+uv'$$

where $u = x$ and $v=(2x+3)^4$

morover $(2x+3)^4$ is a complicated so it is like $(u(\varphi(t)))'\cdot (\varphi(t))'$ which equals to $4(2x+3)^3\cdot 2$

and you have to do it just twice:) (just find derivative of the first derivative)

as A result:

$$\frac{dx}{dy} = (2x+3)^3+8x(2x+3)^3$$

$$\frac{d^2x}{dy} = 6(2x+3)^2+8(2x+3)^3+3\cdot24x(2x+3)^2$$