# Trig differentiation.

#### 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 Last edited by a moderator:

#### AllanCuz

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.

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)

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

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

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

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