How do you differentiate #sin(xy)=ln(x/y)#?

Question

Christopher Agresta · Accepted Answer

The question does not specify the derivative with respect to what? Here are two answers.

#sin(xy)=ln(x/y)#. First, let's rewrite that #ln#. #sin(xy)=ln(x)-ln(y)# Here is the general case first: The derivative with respect to #t# #d/dt(sin(xy))=d/dt(ln(x)) - d/dt(ln(y))# Evaluating the derivatives using the chain rule, we get, #cos(xy)d/dt(xy) = 1/x d/dt(x) - 1/y d/dt(y)#. #cos(xy)[dx/dt y + x dy/dt] = 1/x dx/dt - 1/y dy/dt#. Solve algebraically for whichever one you're interested in. If you want to find the derivative of #y# with respect to #x#, then you're looking for #dy/dx#. You could replace the #t#'s above with #x#'s, but it's probably more clear if we just start over. The derivative with respect to #x# #d/dx(sin(xy))=d/dx(ln(x)) - d/dx(ln(y))# Evaluating the derivatives using the chain rule, we get, #cos(xy)d/dx(xy) = 1/x - 1/y d/dx(y)#. #cos(xy)[y + x dy/dx] = 1/x - 1/y dy/dx#. Before solving for #dy/dx# it is helpful to get rid of the fractions by multiplying both sides of the equation by #xy#. #xycos(xy)[y + x dy/dx] = y - x dy/dx#. #xy^2cos(xy)+x^2ycos(xy) dy/dx = y - x dy/dx#. #x dy/dx + x^2ycos(xy) dy/dx = y - xy^2cos(xy)# #dy/dx = (y - xy^2cos(xy))/(x+x^2ycos(xy))#

Ezekiel Alexander · Answer

undefined To differentiate the given equation $ \sin(xy) = \ln\left(\frac{x}{y}\right) $ with respect to $x$, you use the chain rule and the quotient rule. First, let's differentiate each side of the equation with respect to $x$.

1. Differentiating $ \sin(xy) $ with respect to $x$:

$$ \frac{d}{dx}[\sin(xy)] = \cos(xy) \cdot \frac{d}{dx}[xy] = \cos(xy) \cdot (y + x\frac{dy}{dx}) $$

This step uses the chain rule, and it accounts for $y$ being a function of $x$, hence the term $x\frac{dy}{dx}$.

2. Differentiating $ \ln\left(\frac{x}{y}\right) $ with respect to $x$:

$$ \frac{d}{dx}\left[\ln\left(\frac{x}{y}\right)\right] = \frac{1}{\frac{x}{y}} \cdot \frac{d}{dx}\left[\frac{x}{y}\right] $$

Using the quotient rule for differentiation, $ \frac{d}{dx}\left[\frac{x}{y}\right] = \frac{y(1) - x(\frac{dy}{dx})}{y^2} $, we get:

$$ \frac{d}{dx}\left[\ln\left(\frac{x}{y}\right)\right] = \frac{y - x\frac{dy}{dx}}{xy^2} $$

Setting the derivatives of both sides equal to each other:

$$ \cos(xy) \cdot (y + x\frac{dy}{dx}) = \frac{y - x\frac{dy}{dx}}{xy^2} $$

This equation can then be solved for $\frac{dy}{dx}$ if required, taking into account the relationship between $x$ and $y$.