Let's find the derivative dxdy for the given function.
The function is y=tan(2x)+sin2(3x)+cos2(x).
We will differentiate each term separately using the chain rule.
Step 1: Differentiate the first term, tan(2x).
Let u=2x. Then dxdu=2.
The derivative of tan(u) is sec2(u)⋅dxdu.
dxd(tan(2x))=sec2(2x)⋅2=2sec2(2x)
Step 2: Differentiate the second term, sin2(3x).
This can be written as (sin(3x))2.
Let u=sin(3x). Then dxd(u2)=2udxdu.
Now, find dxdu for u=sin(3x).
Let v=3x. Then dxdv=3.
The derivative of sin(v) is cos(v)⋅dxdv.
So, dxdu=cos(3x)⋅3=3cos(3x).
Now substitute back into the derivative of u2:
dxd(sin2(3x))=2sin(3x)⋅(3cos(3x))=6sin(3x)cos(3x)
Using the identity sin(2θ)=2sinθcosθ, we can simplify this:
6sin(3x)cos(3x)=3⋅(2sin(3x)cos(3x))=3sin(2⋅3x)=3sin(6x)
Step 3: Differentiate the third term, cos2(x).
This can be written as (cos(x))2.
Let w=cos(x). Then dxd(w2)=2wdxdw.
Now, find dxdw for w=cos(x).
dxdw=−sin(x)
Now substitute back into the derivative of w2:
dxd(cos2(x))=2cos(x)⋅(−sin(x))=−2sin(x)cos(x)
Using the identity sin(2θ)=2sinθcosθ, we can simplify this:
−2sin(x)cos(x)=−sin(2x)
Step 4: Combine the derivatives of all terms.
dxdy=dxd(tan(2x))+dxd(sin2(3x))+dxd(cos2(x))
dxdy=2sec2(2x)+3sin(6x)−sin(2x)
The final answer is dxdy=2sec2(2x)+3sin(6x)−sin(2x).
3 done, 2 left today. You're making progress.