Consider the function fxy xy2 fracyx x neq 0 let x gts...

Consider the function $$f(x,y) = xy^2 - \frac{y}{x}, x \neq 0.$$ Let $$x = g(t,s) = \frac{t^2}{s}$$ and $$y = h(t) = \cos t$$. Using the chain rule, find $$\frac{\partial f}{\partial t}$$ and $$\frac{\partial f}{\partial s}$$.

Transcript

00:01 In this problem of chain rule we have given that if x is equals to t sine s so this is t sine s and y is equals to this is t cos s.

00:16 Then we have to find the value of dow 2 that is two time differentiation first this is with respect to x and then with respect to t of the function f of xy.

00:28 So here we can say that z is a function of u and v where u is equals to this is t -s -s -s -s and then v v is equal to t -s -cine cos x so here we have another variable say x and y so we can replace it with x and y also so here x and y and here x is equal to t -s -s and v and y is equals to t causes now first we have to find the differentiation of the function f of xy with respect to s so here we can say this is z z is the function of x and y and x is the function of t and s and y is the function of again t and s so first we are differentiating with respect to s or this is equal to d z divide with d s so this will this would be here we have two ways to reach s so here this would be thou z that means differentiation of z with respect to x and then we have to differentiate x with respect to s plus differentiation of z with respect to y plus differentiation of y with respect to s so we have to reach s so we have one way here and another way is here and now we have to find the differentiation of x with respect to s so here differentiation of x with respect to s so here to s and here we have x is equal to t sin s so this would be here simply t cos s so this is t coss s and this is t cos s actually so this is t cos s and differentiation of here y with respect to s so here we have to again differentiated so this would be minus t sine so this is minus t sin s so this is sin s so this term can be written as differentiation of z with respect to s is equal to this value is t cost so this is simply t cost s multiplied with differentiation of z with respect to x and here plus and this would be with minus and we have to keep it with minus sign so we have to keep it with minus s so minus t and sine s so this is sine s and multiplied with z with respect to y so here we are writing it as differentiation of z with respect to x so here we are writing it as differentiation of z with respect to x is equals to say f1 and differentiation of z with respect to y is say this is f2 so this term can be written as differentiation of z with respect to s is equals to t multiplied with cos s and this is f1 minus t sine s multiplied with f2 now we have to differentiate with respect to t...

Question

Consider the function $$f(x,y) = xy^2 - \frac{y}{x}, x \neq 0.$$ Let $$x = g(t,s) = \frac{t^2}{s}$$ and $$y = h(t) = \cos t$$. Using the chain rule, find $$\frac{\partial f}{\partial t}$$ and $$\frac{\partial f}{\partial s}$$.

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