Compute the derivative dy/dx if 3x*sinx+3yln(y/x)=2e.Express your solution in terms of x and y.

Answer 1

3x*sin(x) + 3yln(y/x) = 2e

Take the derivative of each piece with respect to "x":

Term #1: 3x*sin(x) - use the product rule (u•v)' = u'v + uv' where u = 3x, u' = 3, v = sin(x), v' = cos(x) so its derivative is: 3sin(x) +3xcos(x)

Term #2: 3yln(y/x) - use the product rule (u•v)' = u'v + uv' where u = 3y, u' = 3y', v = ln(y/x), v' = 1/(y/x)•(y/x)' To finish v' this we need to use the quotient rule (u/v)' = (u'v - uv')/v² where u = y, u' = y', v = x, v' = 1, v² = x² so v' = (y'•x - y•1)/x² so putting this into the initial product rule we get:

(3yln(y/x))' = 3y'•ln(y/x) + 3y•(x/y)((y'•x - y)/x² = 3y'ln(y/x) + 3(y'x - y)/x

Term #3: 2e - the derivative of a constant term is zero

Putting these together:

3sin(x) + 3xcos(x) + 3y'ln(y/x) + 3(y'x - y)/x = 0

sin(x) + xcos(x) + y'ln(y/x) + (y'x - y)/x = 0

y'ln(y/x) + (y'x - y)/x = -sin(x) - xcos(x)

y'ln(y/x) + y'x/x - y/x = -sin(x) - xcos(x)

y'ln(y/x) + y' - y/x = -sin(x) - xcos(x)

y'(ln(y/x) + 1) = -sin(x) - xcos(x) + y/x

y' = (y/x - sin(x) - xcos(x))/(ln(y/x) + 1)

I guess you could simplify the numerator with a common denominator but I'm going to leave well enough alone.