Calculus/Derivative

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Expert: - 5/19/2009


Question
Dear Scotto,

what is the procedure to find dy/dx from this ecuation: x=y ln xy

Thanks.

Answer
Implicit differentiation is the only way that I see.

The right side is a produc rule,
the second funtion is a function of a product rule.
Let me explain.

The derivative of x is 1, the derivative of y is y'.

The derivative of xy is a product rule.  It is y + x'y.
This makes the derivative of ln(xy) into ln'(xy) = (y + y'x)/xy.

The last two make the derivative of y ln(xy) into
y'ln(xy) + y ln'(xy), and we just found ln'(xy).

Putting it in gives y' ln(xy) + y(y + y'x)/xy.

Multiply it out, giving y'ln(xy) + y²/xy + y'x/xy.

Cancelling what cancels gives y'ln(xy) + y/x + y'/y.

Remembering the left side of the equations derivative was 1,
lets put that back in, giving 1 = y'ln(xy) + y/x + y'/y.

If we subtract off y/x from both sides, we get
1 - y/x = y'ln(xy) + y'/y.

We can then factor out y' on the right side, giving
1 - y/x = y'(ln(xy) + y).

Now if we divide by the coefficient on the y', we get
(1 - y/x)/(ln(xy) + y) = y'.

To put it in proper format, that's really
y' = (1 - y/x)/(ln(xy) + y).

Calculus

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