Derivative of a trig function f(x) = sin(5x) / cos(x^2)

[pope4]

I'm doing a trig calc question right now and I'm confused as to why I'm not getting the right answer. The question is asking to take the derivative of this equation:

$$f(x)= \frac{sin(5x)}{cos(x^2)}\\$$
I thought I'd get the right answer by applying the quotient rule, and so it led me to an output as follows:

$$\frac{d}{dx}\frac{sin(5x)}{cos(x^2)}=\\ \frac{5(cos(5x))(cos(x^2))-2x(-sin(x^2))(sin(5x))}{cos^2(x^2)}$$
However, I checked my answer with WolframAlpha and that is definitively not what it's supposed to be. I checked the steps for solving it, and it said that I should change it using trig identities so that it's not a fraction. Is there a rule saying I shouldn't take derivatives of equations where 2 trig functions are both the numerator and the denominator? Is there a way to solve it when it's in its original form and, if so, where did I go wrong?

Just as a reference, this is what Wolfram produced:

$$\frac{d}{dx}\frac{sin(5x)}{cos(x^2)}=\\ sin(x^2)(2xsin(5x)tan(x^2)+5cos(5x))$$
Thank you for the help!

 

[khansaheb]

You have the same answer as WA. Simplify your answer and you'll see it.

 

[Steven G]

Did you bother to see if the answers are the same? Using W/A, just compute the difference between the two solutions. If you get 0, then the two solutions are equivalent--now show that they are the same. If the difference is not 0, then at least one of the solutions is wrong.

 

[pope4]

Just as a reference, this is what Wolfram produced:

$$\frac{d}{dx}\frac{sin(5x)}{cos(x^2)}=\\ sin(x^2)(2xsin(5x)tan(x^2)+5cos(5x))$$

I wrote it down wrong, the first term should be sec(x^2)(2xsin(5x)...)

 

[Steven G]

I checked the two 'solutions' when x=0 and they disagreed with one another. At least one of the two solutions is wrong!

 

[Steven G]

I wrote it down wrong, the first term should be sec(x^2)(2xsin(5x)...)

$$\frac{5(cos(5x))(cos(x^2))-2x(-sin(x^2))(sin(5x))}{cos^2(x^2)}$$Divide each term in the numerator by $$cos(x^2)$$ leaving just cos(x^2) in the denominator which you then bring .....

 

[pope4]

You have the same answer as WA. Simplify your answer and you'll see it.

$$\frac{5(cos(5x))(cos(x^2))-2x(-sin(x^2))(sin(5x))}{cos^2(x^2)}$$Divide each term in the numerator by $$cos(x^2)$$ leaving just cos(x^2) in the denominator which you then bring .....

I checked with both WA and through my calculator, they are indeed equal. I did go back and check while I was doing the question, but I must've typed something in wrong because I didn't notice they were the same. Thank you for pointing that out!

 

[Steven G]

I wrote it down wrong, the first term should be sec(x^2)(2xsin(5x)...)

Expressions that are being multiplied are called factors. Expressions that are being added or subtracted are called terms.
I wrote it down wrong, the first term should be sec(x^2)(2xsin(5x)...). It should say that the first factor should be sec(x^2)

 

[pope4]

Expressions that are being multiplied are called factors. Expressions that are being added or subtracted are called terms.
I wrote it down wrong, the first term should be sec(x^2)(2xsin(5x)...). It should say that the first factor should be sec(x^2)

That's true, sorry!

 

[Steven G]

That's true, sorry!

You don't have to be sorry. I prefer that you just learn those definitions.