# Differentiation of Log(cos(X)/x^2)^2

• Anne5632
Thanks for catching that mistake. In summary, the conversation involves differentiating the expression y = (log(t))^2 using the chain rule, derivative of log, and quotient rule. There is a discussion about the correct way to differentiate and simplify the equation, with one approach using the chain rule and the other using the rules of logs.

#### Anne5632

Homework Statement
Differentiate
Relevant Equations
Log(cos(X)/x^2)^2
Im going by the chain rule.
I let y=log(t)^2.
T=cos^2x/x^2

Dy/DT is 2/t * log(t)
Dt/DX is (sin(2x)/X )((sinx+cosx)/X)
Can someone verify this is the correct way ? As when I multiply dydt by dtdx I get an equation I don't know how to simplify

Never mind solved

Anne5632 said:
Homework Statement:: Differentiate
Relevant Equations:: Log(cos(X)/x^2)^2
This expression (it's not an equation) is somewhat ambiguous. Just to verify, is this what you're working with?
$$y = \left[\log\left( \frac {\cos(x)}{x^2}\right)\right]^2$$
Anne5632 said:
Im going by the chain rule.
I let y=log(t)^2.
T=cos^2x/x^2
You need to use more than just the chain rule. I used, in this order, the chain rule, derivative of log, and quotient rule.
Anne5632 said:
Dy/DT is 2/t * log(t)
(Edited to correct my error)
No. If ##y = (\log(t))^2##, then ##\frac{dy}{dt} = 2\cdot \log(t) \frac d{dt} \log(t) = 2\frac {\log(t)} t##
Also, try to be more consistent in your use of variables. You have x, X, t, T, Dy/DT, dydt, dtdx, Dt/DX.
Anne5632 said:
Dt/DX is (sin(2x)/X )((sinx+cosx)/X)
Can someone verify this is the correct way ? As when I multiply dydt by dtdx I get an equation I don't know how to simplify
I don't get anything close to this.

Last edited:
Anne5632
I would start from $$(\log(\cos(x)/x^2))^2 = (\log(\cos(x)) - 2 \log x)^2$$ and use $$\frac{d}{dx} g(x)^2 = 2g(x) \frac{dg}{dx}.$$

vela and Anne5632
Yea instead of making it hard and doing chain in the beginning I simplified with rules of logs

berkeman
Anne5632 said:
Yea instead of making it hard and doing chain in the beginning I simplified with rules of logs.
You, of course, should be able to do it both ways, and it can be good practice to see how to transform/simplify one answer into the other.

Mark44 said:
No. If ##y = (\log(t))^2##, then ##\frac{dy}{dt} = 2\cdot \frac d{dt} \log(t) = \frac 2 t##
@Anne5632 had it right. You're still going to have a ##(\log t)^1## when you apply the chain rule.

Mark44 said:
I don't get anything close to this.
Neither did I.

Mark44 said:
No. If ##y = (\log(t))^2##, then ##\frac{dy}{dt} = 2\cdot \frac d{dt} \log(t) = \frac 2 t##
vela said:
@Anne5632 had it right. You're still going to have a ##(\log t)^1## when you apply the chain rule.
You're right. It should have been ##2\cdot \log(t) \cdot \frac d{dt} \log(t) = 2 \frac{\log(t)} t##.