Differentiating 3 terms with product rule

[fernanhen]

Homework Statement

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

Homework 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.

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.

 

[Infinitum]

Hello feranhen!

Homework Equations[/h2]

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.

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

 

[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?

Thanks for the reply :-)

 

[Infinitum]

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.

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?

How do I write x^2 the way you did by the way?

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.

 

[fernanhen]

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?

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.

 

[Infinitum]

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.

Edit : Yep, its right.

See post below.

 

[HallsofIvy]

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).

 

[Infinitum]

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).

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

 

[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.