Let f(x,y) = xy^2 – x^2, where x = t^3 and y = 3t^2 | Chain Rule & Parametric Differentiation Explained | Query Point Official
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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`
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Frequently Asked Questions (FAQs)
What is the chain rule in calculus?
The chain rule is a differentiation technique used when a function depends on another variable that is itself a function of a different variable.
What is parametric differentiation?
Parametric differentiation is used when variables like x and y are expressed in terms of a third variable, usually denoted by t.
Why is the chain rule used in this problem?
The chain rule is used because both x and y depend on the parameter t, and the function f(x, y) must be differentiated with respect to t.
Is this question important for exams?
Yes, this type of question is very common in university-level exams such as BS Mathematics, Engineering Mathematics, and Calculus courses.
What is the final answer of df/dt?
The final derivative of the given function is:
df/dt = 63t6 + 24t7
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