Question

if u=x2y3, x=log t ,y=et find du/dt​

Answer 1

Answer:

$2\cdot logt\cdot \frac{1}{t}\cdot e^{3t}+3\cdot e^{3t}\cdot log^{2}t$

Step-by-step explanation:

As per the question,

We have provided:

$u =x^{2}y^{3}$

$x =logt$

$y =e^{t}$

Now,

Substitue the value of x and y in the equation of u, we get

$u =log^{2}t\cdot e^{3t}$

Now on differentiating with respect to t by applying the u-v rule:

$\frac{d(u\cdot v)}{dx}=u\cdot \frac{dv}{dx}+v\cdot \frac{du}{dx}$

Therefore,

$u = e^{3t} \cdot log^{2}t$

$\frac{du}{dt}=2\cdot logt\cdot \frac{1}{t}\cdot e^{3t}+3\cdot e^{3t}\cdot log^{2}t$

Hence, $\frac{du}{dt}=2\cdot logt\cdot \frac{1}{t}\cdot e^{3t}+3\cdot e^{3t}\cdot log^{2}t$

Answer 2

Answer:

$\rm \dfrac{du}{dt}=2e^{3t}\dfrac{\log t}{t}+3e^{3t}\log^2t.$

Step-by-step explanation:

Given:

• $\rm u=x^2y^3.$

• $\rm x=\log t.$
• $\rm y=e^t.$

To find:

$\rm \dfrac{du}{dt}.$

We have,

$\rm \dfrac{du}{dt}=\dfrac{d}{dt}\left ( x^2y^3 \right )\\ = y^3\dfrac{d}{dt}\left ( x^2 \right )+x^2\dfrac{d}{dt}\left ( y^3 \right )\\= y^32x\dfrac{dx}{dt}+x^23y^2\dfrac{dy}{dt}\\= 2xy^3\dfrac{dx}{dt}+3x^2y^2\dfrac{dy}{dt}.\ \ \ \ \ \ \ \ ................\ \ eq. (1).$

Also,

$\rm \dfrac{dx}{dt}=\dfrac{d}{dt}\left (\log t \right )=\dfrac 1t.\\\\ \dfrac{dy}{dt}=\dfrac{d}{dt}\left (e^t \right )=e^t.$

Putting the values of all the variables in equation (1), we get,

$\rm \dfrac{du}{dt}=2xy^3\cdot \dfrac 1t+3x^2y^2\cdot e^t\\=2\cdot \log t\cdot (e^t)^3\cdot \dfrac 1t+3\cdot (\log t)^2\cdot(e^t)^2\cdot e^t\\=2e^{3t}\dfrac{\log t}{t}+3e^{3t}\log^2t.$