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sabii
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• jhelum
• Pakistan

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At 1:53pm on June 5, 2015, nAzAkAt Ali said…
At 9:57am on May 27, 2015, Heart *+* Beat said…

Differentiate the following function with respect to t.
w= f(x,y) = x2−xy+y2, where x = g(t)=4t3 and y=h(t) = cost
solution:
if w=f(x,y) , x=g(t) and y=h(t) then by chain rule in function of one variable
then…..
dw/dt= dw/dx . dx/dt + dw/dy. dy/dt
Here we have w = x2 – xy + y2 , x= 4t3 ,y=cost
dw/dx= d/dx ( x2 – xy + y2 )
dw/dx = 2x-y+ y2……………(i)
dx/dt = d/dx (4t3)
dx/dt =12t2…………….(ii)
dw/dy = d/dx ( x2 – xy + y2 )
dw/dy = x2 – x + 2y………………..(iii)
dy/dt = d/dx (cost)
dy/dt = - sint…………………..(iv)

At 9:46am on May 27, 2015, Heart *+* Beat said…

Now by the formula….
dw/dt= dw/dx . dx/dt + dw/dy. dy/dt
dw/dt = ( 2x-y+ y2 )( 12t2 ) + (x2 – x + 2y)( - sint )
By putting the values of x and y……..
dw/dt = [( 2(4t3)-cost+ cost2 )( 12t2 ) + ((4t3)2 – 4t3 + 2cost)( - sint )]
dw/dt = [ 8t3-cost+ cost2 )( 12t2 ) + (16t5 – 4t3 + 2cost)( - sint )]
dw/dt = [( 96t5 - 12t2 cost + 12t2 cost2 ) + (-16t5sint +4t3sint -2costsint)]
dw/dt = ( 96t5 - 12t2 cost + 12t2 cost2 -16t5sint +4t3sint -2costsint)

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