# Differentiating 3 terms with product rule

1. May 28, 2012

### fernanhen

1. The problem statement, all variables and given/known data

Find the derivative of (x^2)(sinx)(tanx)

2. Relevant equations

f(x+y)(d/dx)=(x)(y)d/dx+(y)(x)d/dx

That's for differentiating with 2 terms. With 3, I haven't done it yet.

3. The attempt at a solution

x^2(sinx)(sec^2x)+tanx(sinx)2x+x^2(tanx)(cosx)
x^2(sinx/1)(1/cos^2x)+sinx/cosx)(sinx/1)(2x)+x^2(sinx/cosx)(cosx/a)
sinx/cosx(x^2/cos^2x+2xsinx+x^2cosx)

Another attempt was

2xsinxtanx

Which one do you guys think is right. It may be neither of them.

2. May 28, 2012

### Infinitum

Hello feranhen!

That product rule formula seems incorrect, or I might have misinterpreted the way you wrote it.

The product rule is,

$\frac{\mathrm{d} (f(x) g(x))}{\mathrm{d} x} = f(x)g'(x)+ f'(x) g(x)$

Now consider $f(x) = x^2sin(x)$ and $g(x) = tan(x)$ and solve. You will need to use the product rule twice

3. May 28, 2012

### fernanhen

Ok, here we go:

x^2(sinx)(sec^2(x))+2xcosx(tanx)

I didn't understand what you meant by using the product rule twice. How do I write x^2 the way you did by the way?

4. May 28, 2012

### Infinitum

The bolded part is wrong. This is where you use product rule the second time. Remember, you need $f'(x)= \frac{d}{dx}(x^2sin(x))$ What will you get if you use it here?

The mathematical text is called LaTeX. Here is a nice little tutorial to learn from : https://www.physicsforums.com/showthread.php?t=546968 You can alternatively use the x2 button above the reply box which uses HTML, but LaTeX has more options.

5. May 28, 2012

### fernanhen

I believe this is what I did on first trial of this problem. Maybe I didn't understand the suggestion but in essence what I did was to multiply the derivative of one times the two other terms, then ADD the derivative of the second term times the other 2, then ADD AGAIN the derivative of the third term times the other two.

Without simplifying I got:

(x^2)(sinx)(1/cos^2x)+(sinx/cosx)(sinx)(2x)+(x^2)(sinx/cosx)(cosx)

Sorry for not understanding it promptly but I haven't seen this done in our classroom yet, not with trigonometric functions at least. Can you clarify what may be wrong with the above?

Thanks a lot, I appreciate your help.

6. May 28, 2012

### Infinitum

Edit : Yep, its right.

See post below.

Last edited: May 28, 2012
7. May 28, 2012

### HallsofIvy

Staff Emeritus
To differentiate f(x)g(x)h(x) with respect to x, let F(x)= f(x)g(x) so that you are differentiating F(x)h(x). By the product rule, that is F'(x)g(x)+ F(x)g'(x). And, applying the product rule to F(x)= f(x)g(x), F'(x)= f'(x)g(x)+ f(x)g'(x). Putting that into the first formula, (f(x)g(x)h(x))'= (f'(x)g(x)+ f(x)g'(x))h(x)+ (f(x)g(x))h'(x)= f'(x)g(x)h(x)+ f(x)g'(x)h(x)+ f(x)g(x)h'(x).

8. May 28, 2012

### Infinitum

Wow, I cant believe I didn't realize this mentally and made a silly mistake. Thanks for the clarification, HallsofIvy

9. May 28, 2012

### fernanhen

Yep, thanks HallsofIvy, this : f'(x)g(x)h(x)+ f(x)g'(x)h(x)+ f(x)g(x)h'(x) , is what I did.

Thanks to you both for helping me out with this problem.