Ask Me Help Desk - Differentiate: f(t)= ln(t^2√(2-t))

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Differentiate: f(t)= ln(t^2√(2-t))

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

Quote:

Originally Posted by dodo333

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

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

$<br /> \frac {df}{dt} = \frac {df}{dx} \ \frac {dx} {dt}<br />$

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

Quote:

Originally Posted by ebaines

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

$<br /> \frac {df}{dt} = \frac {df}{dx} \ \frac {dx} {dt}<br />$

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.

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