1. ## Sin/Cos Differentiation

Here are a couple of sin cos differentiation questions. My teacher has gone over the rules for a bit but hasn't shown how to do the actual equation questions.

2. Originally Posted by SportfreundeKeaneKent
Here are a couple of sin cos differentiation questions. My teacher has gone over the rules for a bit but hasn't shown how to do the actual equation questions.

f(x) = cos(pi/x)

put g(x)=pi/x, then:

f(x) = cos(g(x))

Now we use the chain rule:

f'(x) = -sin(g(x)) g'(x) = -sin(pi/x) (-pi/x^2)

so f'(2) = sin(pi/2) (pi/4) ~= 0.785

RonL

3. Originally Posted by SportfreundeKeaneKent
Here are a couple of sin cos differentiation questions. My teacher has gone over the rules for a bit but hasn't shown how to do the actual equation questions.

y = x cos(x)/(1+e^x)

we find dy/dx using the product rule and the quotient rule:

dy/dx = d/dx[x/(1+e^x)] cos(x) + [x/(1+e^x)] d/dx cos(x)

Now you need to use the quotient rule to do the first of the derivatives on
the right above. Me I don't know the quotient rule so I will use the product
rule a second time:

d/dx[x/(1+e^x)] = 1/(1+e^x) - x e^x /(1+e^x)^2 = (1 - e^x (x - 1))/(e^x + 1)^2

so:

dy/dx = [(1 - e^x (x - 1))/(e^x + 1)^2] cos(x) - [x/(1+e^x)] sin(x)

RonL

4. How do you differentiate this one since it has both sine and cosine in it?

5. Originally Posted by SportfreundeKeaneKent
How do you differentiate this one since it has both sine and cosine in it?

Individual terms and the chain rule:
d/dx[sin^2(x)] = 2*sin(x)*cos(x)
d/dx[cos(2x - 1)] = -sin(2x - 1) * 2

So
dy/dx = 12*sin(x)*cos(x) - 2sin(2x - 1)

-Dan