Trig differentiation.

May 2010
36
0
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:
Apr 2010
384
153
Canada
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.
 
Dec 2009
3,120
1,342
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)

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

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

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

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