MHB Trig Calculus: Solving y = sin^2(x) / cos^2 (x)

[tmt1]

Hi,

I'm working on this problem:

y = sin^2(x) / cos^2 (x)

what is dy/dx?

In the solutions it says to convert (sin(x)/cos(x))^2=tan^2(x)

And the answer is dy/dx = 2sec^2(x)tanx

However, using the quotient rule, I got this answer:

dy/dx =[ (2sin(x)cos(x))cos^2(x) - (sin^2x)2(cos(x))(-sin(x))] / cos^2(x)

=[2sin(x)cos^2(x)+2sin^3(x)]/cos^2 (x)

Does this mean

[2sin(x)cos^2(x)+2sin^3(x)]/cos^2 (x) = 2sec^2(x)tanx ?

 

[Chris L T521]

Hi,

I'm working on this problem:

y = sin^2(x) / cos^2 (x)

what is dy/dx?

In the solutions it says to convert (sin(x)/cos(x))^2=tan^2(x)

And the answer is dy/dx = 2sec^2(x)tanx

However, using the quotient rule, I got this answer:

dy/dx =[ (2sin(x)cos(x))cos^2(x) - (sin^2x)2(cos(x))(-sin(x))] / cos^2(x)

=[2sin(x)cos^2(x)+2sin^3(x)]/cos^2 (x)

Does this mean

[2sin(x)cos^2(x)+2sin^3(x)]/cos^2 (x) = 2sec^2(x)tanx ?

You made some mistakes with your algebra and applying the formula properly. If you use quotient rule, you should end up with
\[\begin{aligned} \frac{(2\sin x\cos x)\cos^2 x - (\sin^2 x)(2\cos x(-\sin x))}{(\cos^2 x)^2} &= \frac{2\sin x\cos^3x +2\sin^3 x\cos x}{\cos^4x} \\ &= \frac{2\sin x\cos x(\sin^2 x+\cos^2x)}{\cos^4x} \\ &= \frac{2\sin x}{\cos^3 x}\\ &= 2\frac{\sin x}{\cos x}\cdot\frac{1}{\cos^2 x}\\ &= 2\tan x\sec^2 x\end{aligned}\]

I hope this makes sense! (Smile)

 

[Deveno]

You forgot to square the denominator...