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I have to find the derivative of the following function:

$$f(x) = (x^3+ 4)(4x^5 + 2x − 5)^{1/2}$$

To start solving this, I've dissected the equation and realize that I must use the product and chain rule to find the derivative. The problem is I don't know how to use both at the same time. How do I do this? How can I tell when to use what first?

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Try to only focus on one of them at a time. With differentiation, you usually work from the outside inwards, so figure out which part is the most inward so you can do that part last.

It's the power that is telling you that you need to use the chain rule, but that power is only attached to one set of brackets. It's the fact that there are two parts multiplied that tells you you need to use the product rule. Since the power is inside one of those two parts, it is going to be dealt with after the product.

So we'll do the product rule first.

Adding an extra set of brackets should help to know which part is "innermost": $$ f(x) = (x^3 +4)\left[ (4x^5 +2x-5)^\frac12\right] $$

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  • $\begingroup$ This was very helpful, thank you. The only thing now is that the next step will equate to f'(x)=(3x^2)(4x^5 + 2x−5)^1/2 + (x^3 + 4)(1/2)(4x^5 + 2x − 5)^-1/2(20x^4 +2). What happened to the g'[h(x)] when applying the chain rule second? It seems as if h'(x) is there (which is the (20x^4 +2)) but not the other part. $\endgroup$ – Louzzy56 Jul 25 '14 at 7:29
  • $\begingroup$ The $g'[h(x)]$ is the $\frac12(4x^5 +2x-5)^{-\frac12}$. $\endgroup$ – DavidButlerUofA Jul 25 '14 at 7:40
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$$ f(x) = (x^3+4)\sqrt{4x^5 + 2x - 5} = g(x) \cdot h(x) $$ where $$ g(x) =(x^3+4)\\ h(x) = \sqrt{4x^5 + 2x - 5} $$ $$ \frac{df}{dx} = g'(x)h(x) + g(x) h'(x) $$ Now lets take this step by step: $$ g'(x) = 3x^2 $$ Now for $h'(x)$ let $$ u(x) = 4x^5+2x-5 $$ Chain rule yields $$ \frac{dh}{dx} = \frac{dh}{du} \cdot \frac{du}{dx} = \frac{1}{2 \sqrt{u(x)}} \cdot u'(x) $$ where $$ u'(x) = 20x^4+2 $$ so, in summation: $$ \frac{df}{dx} = 3x^2 \sqrt{4x^5+2x-5} + (x^3+4) \cdot \frac{20x^4+2}{2 \sqrt{4x^5+2x-5}} $$ If I didn't make any mistake. So always take one step at a time, finding all the components you need and it shouldn't be too difficult =)

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  • $\begingroup$ Whats with the downvote? $\endgroup$ – Noxet Jul 25 '14 at 7:02
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You have

$$\begin{align} f(x) & = g(x) h(x)^{1/2} \\[1.5ex] f'(x) &= g'(x)h(x)^{1/2} + g(x)(h(x)^{1/2})' & \text{by product rule} \\ &= g'(x)h(x)^{1/2} + g(x)\color{blue}{(\tfrac 12 h(x)^{-1/2} h'(x))} & \text{by chain rule} \end{align}$$

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You have two factors, so the product rule tells you:

$$(h(x)g(x))'=h'(x)g(x)+h(x)g'(x)$$

Then you get:

$$f'(x)=(x^3+4)'(4x^5+2x-5)^{1/2}+(x^3+4)(4x^5+2x-5)'^{1/2}$$

And now you apply the chain rule to get the derivative of the second term.

$$f'(x)=3x^2(4x^5+2x-5)^{1/2}+\frac{(x^3+4)(20x^4+2)}{2(4x^5+2x-5)'^{1/2}}$$

The situation you have is this one:

$$f'(x)=h(x)g(r(x))$$

So your first apply the product rule to separate the factors and the apply the chain rule to get the derivative of $g(r(x))$.

$$f'(x)=h'(x)g(r(x)) + h(x)r'(x)g'(r(x))$$

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Hint:

You have $f(x) = (x^3+ 4)(4x^5 + 2x − 5)^{1/2}=g(x) \cdot \left( h(x) \right) ^{1/2}$

$f'(x)=g'(x) \cdot \left( h(x) \right) ^{1/2}+g(x) \cdot {1/2}\cdot h(x) ^{-1/2} \cdot h'(x)$

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Let $u=x^3+4$ and $v= (4x^5+2x-5)^{1/2}$. Then $$u'= 3x^2 $$ and $$v'= (10x^4+1)(4x^5+2x-5)^{-1/2} $$ So $$f'(x)=3x^2(4x^5+2x-5)^{1/2}+(x^3+4)(10x^4+1)(4x^5+2x-5)^{-1/2}$$

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