# Derivative using Product Rule

I have a derivative:

$$\frac{d}{dx}[\sqrt{10x}]$$

I start by using $$^n\sqrt{a^{x}}=a^\frac{x}{n}$$ to give me:

$$\frac{d}{dx}[(10x)^\frac{1}{2}]$$

I can factor out 10 to give me:

$$\frac{d}{dx}[(10(x))^\frac{1}{2}]$$

This is the part that loses me. I know I need to use the Product Rule to $$10(x)$$, but I must not be setting it up correctly. When I use the Product Rule I get:

$$(f*g)'=f'*g+f*g'$$ $$(f*g)'=(10)'*(x)+(10)*(x)'$$ $$(f*g)'=0*(x)+(10)*1$$ $$(f*g)'=10$$

What I've been told the result should be after using the Product Rule is: $$\frac{d}{dx}[10^\frac{1}{2}x^\frac{1}{2}]$$

Where would I be going wrong when it comes to using the Product Rule?

• Note $g=x^{1/2}$ not $g=x$ as you wrote.
– zwim
Mar 18 at 16:31
• You may use the product rule on $10x$ if you really want to, but you also need to use the chain rule: you have been asked to calculate the derivative of $(10x)^{1/2}$, after all, not of $10x$.
– user239203
Mar 18 at 16:33
• Be careful with your notation! $\mathrm d/\mathrm dx = f(x)$ doesn't make any sense. You could write $y = f(x)$ and then $\mathrm dy/\mathrm dx = \dots$, or you could write $\frac{\mathrm d}{\mathrm dx}(f(x)) = \dots$. Mar 18 at 16:41
• Zak: You´re are supposed to accept an answer, if you have no questions anymore. It also show that you appreciate the work of the people here. Check your other questions as well. Thanks in advance. Mar 19 at 1:24

## 2 Answers

When you are multiplying two functions and one of them is just a constant, then when you take the derivative, you can just pull the constant out. No need for the product rule in that case.

For this problem, you don't need the chain rule either, just the power rule! $$\frac{d}{dx}[10x]^{1/2} = 10^{1/2} \frac{d}{dx} x^{1/2} = 10^{1/2}\cdot \frac{1}{2} \cdot x^{-1/2} = \frac{\sqrt{10}}{2 \sqrt{x}}$$

If you want to use the chain rule instead, then $$10x$$ is your inside function and $$\sqrt{x}$$ is your outside function.

A couple of points: you wrote $$\frac{d}{dx} = [10x]^{1/2}$$ when what you mean is $$\frac{d}{dx} [10x]^{1/2}$$. That is not an equation, it is an expression, the derivative operator, acting on a function.

Also, your final result does not have calculus done, it was just the bit of algebra that says $$[10x]^{1/2} = 10^{1/2} x^{1/2}$$. Maybe you understood that, but your first mistake makes it a little hard to tell.

Since constants have a zero derivative, don't bother applying the product rule to them, just factor them out.

i.e. $$(c\times f(x))' = 0\times f(x)+c\times f'(x)=c\times f'(x)$$ is the same as $$c\times (f(x))'$$ where $$c$$ was factored out.

• by factoring out the constant

$$(\sqrt{10x})'=(\sqrt{10}\times\sqrt{x})'=\sqrt{10}\times (\sqrt{x})'=\sqrt{10}\times (x^\frac 12)'=\sqrt{10}\times \frac 12x^{-\frac 12}=\dfrac{\sqrt{10}}{2\sqrt{x}}$$

• by chain rule

$$\require{cancel}((10x)^\frac 12)'=\frac 12(10x)^{-\frac 12}\times (10x)'=\frac 12(\cancel{10}x)^{-\frac 12}\times \underbrace{10}_{10^\frac 12\cdot\cancel{10^\frac 12}}=\frac 1210^\frac12x^{-\frac12}=\dfrac{\sqrt{10}}{2\sqrt{x}}$$

And this is the same result.