Exponential and Logarithmic Differentiation. Q. If xe^(xy)=y+sin^2 x, then find dy/dx at x=0. If we differentiate the function directly as follows: e^(xy)+xe^(xy)[y+x(dy)/(dx)]=(dy)/(dx)+sin(2x) At x=0 We have 1+0=(dy)/(dx)+0 which gives dy/dx=1 But if we apply logarithm and differentiate we have: log(x)+xy=log(y+sin^2 x) which when differentiated gives (1)/(x)+y+x(dy)/(dx)=((dy)/(dx)+\sin2x)/(y+\sin^2x) If we take x=0 here, we get undefined values. And we are unable to obtain the value of dy/dx individually. Is there any mistake in the logarithmic method?

Brandon Monroe

Brandon Monroe

Answered question

2022-08-12

Exponential and Logarithmic Differentiation.
Q. If x e x y = y + sin 2 x, then find d y d x at x=0.
If we differentiate the function directly as follows:
e x y + x e x y [ y + x d y d x ] = d y d x + sin ( 2 x )
At x=0 We have
1 + 0 = d y d x + 0 which gives
d y d x = 1
But if we apply logarithm and differentiate we have:
log ( x ) + x y = log ( y + sin 2 x ) which when differentiated gives
1 x + y + x d y d x = d y d x + sin 2 x y + sin 2 x
If we take x=0 here, we get undefined values. And we are unable to obtain the value of dy/dx individually. Is there any mistake in the logarithmic method?

Answer & Explanation

peculiopy

peculiopy

Beginner2022-08-13Added 8 answers

If x=0 then log(x) is undefined. If you continue and solve for dy/dx you get the indeterminate form 0/0 as x approaches 0.
sarahkobearab4

sarahkobearab4

Beginner2022-08-14Added 5 answers

if you at the constraint
x e x y = y + sin 2 x
you see that at x = 0 , gives y = 0. now for
y = x +  for  x = 0 +
this implies d y d x = 1 at x = , y = 0.
edit:
even you logarithmic differentiation result gives you the same. here is the reason why. you have
1 x + y + x d y d x = d y d x + sin 2 x y + sin 2 x
if you look at the behavior off y for x = 0 + , we find that
1 x = 1 y +
again y = x + balances the dominant term 1 x if we take d y d x = 1 +

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