1. ## Trig differentiation.

Hi,
I'm having a lot of difficulty differentiating the following trig problems:

Question : Find $\displaystyle \frac{dy}{dx}$ for the following functions(Do not simplify the answer).

a) $\displaystyle 4 cos^2(\pi x)$

b) $\displaystyle tan^2 (cos x)$

c)$\displaystyle sin x +sin y=1$

I'm not even sure about how to start on these problems, and would greatly appreciate any helpful tips.

Thanks

2. Originally Posted by spoc21
Hi,
I'm having a lot of difficulty differentiating the following trig problems:

Question : Find $\displaystyle \frac{dy}{dx}$ for the following functions(Do not simplify the answer).

a) $\displaystyle 4 cos^2(\pi x)$

b) $\displaystyle tan^2 (cos x)$

c)$\displaystyle sin x +sin y=1$

I'm not even sure about how to start on these problems, and would greatly appreciate any helpful tips.

Thanks
Well,

1: $\displaystyle y = 4 cos^2(\pi x)$

$\displaystyle = 4 cos( \pi x ) cos( \pi x )$

Then,

$\displaystyle \frac{dy}{dx} = 4[- \pi sin( \pi x) cos( \pi x) - \pi sin( \pi x) cos ( \pi x ) ]$

3: $\displaystyle sin x +sin y=1$

$\displaystyle cosx + cosy y^{ \prime } = 0$

$\displaystyle y^{ \prime} = - \frac{cosx}{cosy}$

Compute 2 in the same way as 1.

3. Originally Posted by spoc21
Hi,
I'm having a lot of difficulty differentiating the following trig problems:

Question : Find $\displaystyle \frac{dy}{dx}$ for the following functions(Do not simplify the answer).

a) $\displaystyle 4 cos^2(\pi x)$

b) $\displaystyle tan^2 (cos x)$

c)$\displaystyle sin x +sin y=1$

I'm not even sure about how to start on these problems, and would greatly appreciate any helpful tips.

Thanks
a)

$\displaystyle y=4cos^2({\pi}x)$

$\displaystyle u={\pi}x$

$\displaystyle v=cosu$

$\displaystyle w=v^2$

$\displaystyle y=4w$

By the chain rule...

$\displaystyle \frac{dy}{dx}=\frac{dy}{dw}\ \frac{dw}{dv}\ \frac{dv}{du}\ \frac{du}{dx}$

b)

similarly

c)

$\displaystyle \frac{d}{dx}\ sinx+\frac{d}{dx}siny=\frac{d}{dx}1=0$

$\displaystyle cosx+\frac{dy}{dx}\ \frac{d}{dy}siny=0$

$\displaystyle \frac{dy}{dx}(cosy)=-cosx$

$\displaystyle \frac{dy}{dx}=-\frac{cosx}{cosy}$