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

[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^2Dy/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

 

[Anne5632]

Never mind solved

 

[Mark44]

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?
$$y = \left[\log\left( \frac {\cos(x)}{x^2}\right)\right]^2$$

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.

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.

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.

 

[pasmith]

I would start from <br /> (\log(\cos(x)/x^2))^2 = (\log(\cos(x)) - 2 \log x)^2 and use <br /> \frac{d}{dx} g(x)^2 = 2g(x) \frac{dg}{dx}.

 

[Anne5632]

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

 

[vela]

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.

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.

I don't get anything close to this.

Neither did I.

 

[Mark44]

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.

You're right. It should have been $2\cdot \log(t) \cdot \frac d{dt} \log(t) = 2 \frac{\log(t)} t$.