Help with derivative of trigonometric functions

[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.

 

[Chestermiller]

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.

 

[SammyS]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

[ b]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} [/b]
I realize how much of a mess it is. I suspect it's something very simple that I missed.

Hello Splice1108. Welcome to PF !

Please don't display your posts in bold typeface.

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))​

Added in Edit:
I see that you have corrected the boldface type.

 

[Mark44]

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

First off, you don't find y^I 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.

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.

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.

 

[Chestermiller]

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.

 

[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.

 

[Dick]

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.

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!