Let f(x,y) = xy^2 – x^2, where x = t^3 and y = 3t^2

Let  `f(x,y) = xy^2 – x^2`, where `x = t^3` and `y = 3t^2` . Use chain rule to find `\frac{df}{dt}` and express the answer in variable t only

Solution:

Here

 `f(x,y) = xy^2 – x^2 ,                         x = t^3 and y = 3t^2`

Now we have to find                    

`\frac{df}{dt}` in variable t only

We know that

`\frac{df}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} -  -  -  -  -  -  -  -  - (1)`

`\frac{dx}{dt} = \frac{d}{dt}(t^3) = 3t^2 -  -  -  -  - (i)`

`\frac{dy}{dt} = \frac{d}{dt}(3t^2) = 6t -  -  -  -  - (ii)`

`\frac{\partial f}{\partial x} = \frac{\partial }{\partial x}f(x,y) = \frac{\partial }{\partial x}(xy^2 - x^2)`

`= \frac{\partial }{\partial x}(xy^2) - \frac{\partial }{\partial x}(x^2) = y^2 - 2x -  -  -  -  - (iii)`

`\frac{\partial f}{\partial y} = \frac{\partial }{\partial y}f(x,y\right) = \frac{\partial }{\partial y}(xy^2 - x^2)` 

`= \frac{\partial }{\partial y}(xy^2) - \frac{\partial }{\partial y}(x^2) = 2xy - 0 = 2xy -  -  -  - (iv)`

From equation (i),(ii),(iii) and (iv), (1) becomes

`\frac{df}{dt} = (y^2 - 2x)(3t^2)(2xy)(6t)`

Now by putting the value of x and y,

`\frac{df}{dt} = ((3t^2)^2 - 2(t^3))(3t^2) + (2(t^3)(3t^2))(6t)`

`\frac{df}{dt} = (3t^2+2 - 2t^3)(3t^2) + ({6t^{3 + 2})(6t)`

`\frac{df}{dt} = (3t^4 - 2t^3)(3t^2) + (6t^5)(6t)`

`\frac{df}{dt} = (9t^{4 + 2} - 6t^{3 + 2}) + (36t^{5 + 1})`

`\frac{df}{dt} = (9t^6 - 6t^5) + (36t^6)`

`\frac{df}{dt} = 9t^6 - 6t^5 + 36t^6`

`\frac{df}{dt} = 45t^6 - 6t^5`