# Derivative help#2

## 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?

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

$$= \frac{sin2x( 1+ctgx) +1 }{(1+ ctgx)^2} +\frac{-sin2x(1+tgx)-1}{(1+ tgx)^2}$$

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

Cyosis
Homework Helper
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.

Last edited:
Mark44
Mentor
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.

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

Mark44
Mentor
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.