Derivative of Trigonometric Expression: y' = -cos2x | Derivative Help #2

[Lynne]

Homework Statement

y= \frac{sin^2 x}{1+ ctgx} +\frac{cos^2 x}{1+ ctgx}

Find: y'

The Attempt at a Solution

Answer is-cos2x but again I can't find it.

Where am I making a mistake?

y'=(\ \frac{sin^2 x}{1+ ctgx} +\frac{cos^2 x}{1+ ctgx} \ )'= \frac{(sin^2 x)'(1+ ctgx) -( sin^2 x) (1+ ctgx)'}{(1+ ctgx)^2}\ + \ \frac{(cos^2 x)'(1+ tgx) - cos^2 x (1+ tgx)'}{(1+ tgx)^2} =


<br /> = \frac{sin2x( 1+ctgx) +1 }{(1+ ctgx)^2} +\frac{-sin2x(1+tgx)-1}{(1+ tgx)^2}<br />

 

[nicksauce]

Well the denominator in the second term goes from being 1 + ctgx to being 1 + tgx.

 

[Cyosis]

The only mistake you've made so far is the function y you gave us. The denominator of the second term should have a tangent, not a cotangent. You have done this correctly during your differentiation. I am afraid it's just a matter of simplification now by using trig identities.

 

[Mark44]

Concerning the problem you posted (which is apparently not the right problem), for future reference you could have made life easier by doing some simplification before doing the differentiation.
y= \frac{sin^2 x}{1+ ctgx} +\frac{cos^2 x}{1+ ctgx}
\Leftrightarrow y = \frac{sin^2 x + cos^2 x}{1+ ctgx}
\Leftrightarrow y = \frac{1}{1+ ctgx}
Now you have a much simpler expression to differentiate, which makes it less likely that you'll make errors.

 

[Lynne]

Yes, it's my mistake. Denominator of the second term is tangent.

 

[Mark44]

Yes, it's my mistake. Denominator of the second term is tangent.

Right, I got that. The point of my post was that you made the (incorrect) problem harder than you needed to by not simplifying first. Always try to do things the simplest way you can get away with.