Please differentiate: f(t)= ln(t^2√(2-t)) using dt/dx

Please differentiate: f(t)= ln(t^2√(2-t)) using dt/dx

I'm guessing that you are alluding to the chain rule:


\frac {df}{dt} = \frac {df}{dx} \ \frac {dx} {dt}


Here let x = t^2 \sqrt{2-t}, so that f = ln(x). Now apply the chain rule - what do you get?

I'm guessing that you are alluding to the chain rule:


\frac {df}{dt} = \frac {df}{dx} \ \frac {dx} {dt}


Here let x = t^2 \sqrt{2-t}, so that f = ln(x). Now apply the chain rule - what do you get?

Hello, thanks for your answer, but I really do not know how do this, I just know that the answer must be obtained using dt/dx. Please let me know the solution with the steps, I would really appreciate that.

I won't give you the answer, but I'll show you how to do a similar problem. Let me know if there are any steps here that you don't understand:

Differentiate: f(t) = sin(t^3 \sqrt t )

1. Let x = t^3 \sqrt t.

2. f = \sin (x), hence \frac {df}{dx} = \cos(x) = \cos(t^3 \sqrt t)

3. \frac {dx}{dt} = \frac {d( t^3 \sqrt t )}{dt} = 3t^2 \sqrt t - \frac {t^3} {2\sqrt t}

4. \frac {df}{dt} = \frac {df}{dx} \ \frac {dx}{dt} = \cos(t^3 \sqrt t ) (3t^2 \sqrt t - \frac {t^3} {2\sqrt t} )

You can apply this same technique to your problem. Post back with what you get for an answer.