# Help with derivative of trigonometric functions

1. Feb 28, 2013

### Splice1108

Find y$^{I}$ (sinxsecx)/1+xtanx The supplied answer is 1/(1+xtanx)$^{2}$

I got stuck with an extra x on top at the end. Where did I mess up at?

y$^{I}$(sinxsecx)/1+xtanx = [1+xtanx*f$^{I}$(sinxsecx)-sinxsecx*f$^{I}$(1+xtanx)]/(1+xtanx)$^{2}$ = [1+xtanx((cos/1)*(1/cos))-((cos/1)*(1/cos))*(0-1*(-cos/sin))]/(1xtanx)$^{2}$=[1+x*(sinx/cosx)*(cosx/sinx)*(-1)+1]/(1+xtanx)$^{2}$=(1+(x)(1)(-1)+1)/(1+xtanx)$^{2}$
I realize how much of a mess it is. I suspect it's something very simple that I missed.

Last edited by a moderator: Feb 28, 2013
2. Feb 28, 2013

### Staff: Mentor

Begin by manipulating your initial expression first:

$$\frac{\sin x\sec x}{(1+x\tan x)}=\frac{\frac{\sin x}{\cos x}}{(1+x\tan x)}=\frac{\tan x}{(1+x\tan x)}=\frac{1}{(x+\cot x)}$$
This should be much easier to differentiate.

3. Feb 28, 2013

### SammyS

Staff Emeritus
Hello Splice1108. Welcome to PF !

Please DO leave the supplied text from the template in tact. -- That should remain in bold.

Parentheses (and other symbols) are important.

I assume your problem is something like:
Find y' for the following.

y = (sin(x)sec(x))/(1 + x∙tan(x))

I see that you have corrected the boldface type.

4. Feb 28, 2013

### Staff: Mentor

First off, you don't find yI or y' of something. y' already is the derivative of y. To indicate that you intend to take the derivative, but haven't done so yet, use the d/dx operator, as in this example:
d/dx(x2) = 2x.
And it's very difficult to read.

This is what you want to find:
$$\frac{d}{dx} \frac{sin(x) sec(x)}{1 + x tan(x)}$$

The first thing to do, since this is a quotient, is to use the quotient rule, so
$$\frac{d}{dx} \frac{sin(x) sec(x)}{1 + x tan(x)} = \frac{(1 + xtan(x))d/dx(sin(x) * sec(x)) - sin(x) * sec(x) * d/dx(1 + x tan(x))}{(1 + x tan(x))^2}$$

Keep working at it a little at a time.

5. Feb 28, 2013

### Staff: Mentor

Have you guys tried taking the derivative of the final relationship I manipulated the initial function into in post #2? All the messy math is removed when you start out with this relationship. It only took me a minute or two to get the final answer. Try it an see.

6. Feb 28, 2013

### Splice1108

I'm going to try to retype my attempted work in a more readable manner.

I used the quotient rule to get:
$\frac{(1+xtanx)\frac{d}{dx}(sinxsecx)-(sinxsecx)\ast\frac{d}{dx}(1+xtanx)}{(1+xtanx)}$

Then went on trying to cancel out the top since the denominator was already the value i was trying to achieve.
$\frac{(1+xtanx)(-cosx\ast secx\ast tanx)-(sinxsecx)\frac{d}{dx}(1+xtanx)}{(1+xtanx)^{2}}$

I'm unsure how to go about differentiating the (1+xtanx) Is it just (sec$^{2}$x)?

I'm stumped as far as where the next step should be.

Also, I appreciate the assistance with getting started on here.

7. Feb 28, 2013

### Dick

No, the derivative of (1+x*tan(x)) isn't sec(x)^2. You need to start using the product rule. Same issue for the first derivative. That is a big step up in readability though!

Last edited: Feb 28, 2013