dy/dt for ln x =e^y+(x^2+y^3)^2-2 if x=1 y=0 and dx/dt=-2
Hi, I am pleased to do this question for you. I am giving the detailed solution below.
ln x =e^y+(x^2+y^3)^2-2
differentiating this with respect to dt
d(ln x)/dt = d(e^y)/dt + d((x^2+y^3)^2) - d(2)/dt
(1/x)dx/dt = (e^y)dy/dt + 2*((x^2+y^3)(2x*dx/dt + (3*y^2)dy/dt) - 0 - > (equation 1)
differential of ln x is 1/x
differential of e^y is e^y but we are doing it with respect to dt so dy/dt term will come
similarly for the next term and differential of constant(2) is always 0.
putting the given values in equation 1,
(1/1)(-2) = (e^0)dy/dt + 2*((1^2 + 0)(2*1*-2 + 0))
-2 = dy/dt + 2*-4
dy/dt = 6 ---------> ANSWER
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