cos derivative problem

[Jason76]

\(\displaystyle y = a^{3} + \cos^{2} x\)

\(\displaystyle y' = 3a^{2} + [u^{2} du]\)

\(\displaystyle y' = 3a^{2} + [u^{2}(-\sin x)]\)

\(\displaystyle y' = 3a^{2} + [-\cos^{2} x \sin x]\)

\(\displaystyle y' = 3a^{2} -\cos^{2} x \sin x\) computer homework says this is wrong. What went wrong?

 

[pka]

\(\displaystyle y = a^{3} + \cos^{2} x\)

\(\displaystyle y = a^{3} + \cos^{2} x\)

\(\displaystyle y' =2\cos(x)[-\sin(x)]=-2\cos(x)\sin(x)\)

\(\displaystyle y = a^{3}\) is a constant, As such its derivative is \(\displaystyle 0\).

 

[Jason76]

\(\displaystyle y = a^{3} + \cos^{2} x\)

\(\displaystyle y' =2\cos(x)[-\sin(x)]=-2\cos(x)\sin(x)\)

\(\displaystyle y = a^{3}\) is a constant, As such its derivative is \(\displaystyle 0\).

So simple. Man, why didn't I see that?

 

[Deleted member 4993]

I hope, you have noticed that you had differentiated cos²(x) incorrectly.

 

[Jason76]

I hope, you have noticed that you had differentiated cos²(x) incorrectly.

This was correct on homework:

\(\displaystyle y = a^{3} + \cos^{2} x\)

\(\displaystyle y' = [u^{2} du]\)

\(\displaystyle y' = [u^{2}(-\sin x)]\)

\(\displaystyle y' = [-\cos^{2} x \sin x]\)

\(\displaystyle y' = -\cos^{2} x \sin x\)

 

[lookagain]

This was correct on homework:

\(\displaystyle y = a^{3} + \cos^{2} x\)

\(\displaystyle y' = [u^{2} du]\)

\(\displaystyle y' = [u^{2}(-\sin x)]\)

\(\displaystyle y' = [-\cos^{2} x \sin x]\)

\(\displaystyle y' = -\cos^{2} x \sin x \ \ \ \ \ \ \ \)<------- * If * this were the y prime, then y would have been equal to \(\displaystyle \ \ \dfrac{1}{3}\cos^3(x) \ + \ C.\)

.