Use implicit differentiation to show
dy/dx = [(1-xy)cot(y)]/[x^2 cot(y) + xcsc^2(y)]
if xy = ln(x cot y).
from xy = ln(x coty)
x dy/dx + y = (x(-csc^y dy/dx) + coty)/(x coty)
cross-multiply
(x^2)(coty)dy/dx + xycoty = -x(csc^2y)dy/dx + coty
(x^2)(coty)dy/dx + x(csc^2y)dy/dx = coty - xycoty
dy/dx((x^2)(coty) + x(csc^2y)) = coty(1 - xy)
dy/dx = (1-xy)coty/[(x^2)(coty)dy/dx + x(csc^2y)]
To find the derivative of \(y\) with respect to \(x\) using implicit differentiation, start with the given equation:
\[xy = \ln(x \cot y)\]
We will differentiate both sides of the equation with respect to \(x\).
Differentiating \(xy\) with respect to \(x\) gives:
\[\frac{d}{dx} (xy) = \frac{d}{dx} (\ln(x \cot y))\]
Applying the product rule on the left side:
\[y + x \frac{dy}{dx} = \frac{d}{dx} (\ln(x \cot y))\]
Now, let's find the derivative of \(\ln(x \cot y)\). We use the chain rule here:
\[\frac{d}{dx} (\ln(x \cot y)) = \frac{1}{x \cot y} \cdot \frac{d}{dx} (x \cot y)\]
Simplifying the expression further:
\[\frac{d}{dx} (\ln(x \cot y)) = \frac{\cot y}{x} \cdot \frac{d}{dx} (x) + x \cdot \frac{d}{dx} (\cot y)\]
Using the quotient rule to find \(\frac{d}{dx} (\cot y)\):
\[\frac{d}{dx} (\cot y) = \frac{(-\csc^2 y) \cdot \frac{dy}{dx}}{1 + \cot^2 y}\]
Substituting the above derivative and simplifying, we have:
\[\frac{d}{dx} (\ln(x \cot y)) = \frac{\cot y}{x} + \frac{-x \csc^2 y \frac{dy}{dx}}{1 + \cot^2 y}\]
Bringing it back to our original equation:
\[y + x \frac{dy}{dx} = \frac{\cot y}{x} + \frac{-x \csc^2 y \frac{dy}{dx}}{1 + \cot^2 y}\]
Now, let's rearrange the terms to isolate \(\frac{dy}{dx}\) and simplify the expression:
\[x \frac{dy}{dx} + x \csc^2 y \frac{dy}{dx} = \frac{\cot y}{x} - y\]
Combining like terms and factoring out \(\frac{dy}{dx}\):
\[x \left(1 + \csc^2 y\right) \frac{dy}{dx} = \frac{\cot y - xy}{x}\]
Simplifying the expression further:
\[\frac{dy}{dx} = \frac{\cot y - xy}{x \left(1 + \csc^2 y\right)}\]
Finally, using trigonometric identities:
\[\frac{dy}{dx} = \frac{(1 - xy) \cot y}{x^2 \cot y + x \csc^2 y}\]
Therefore, we have shown that:
\[\frac{dy}{dx} = \frac{(1 - xy) \cot y}{x^2 \cot y + x \csc^2 y}\]
To find the derivative using implicit differentiation, let's start by differentiating both sides of the given equation with respect to x.
xy = ln(x cot y)
To differentiate xy, we use the product rule. The derivative of xy with respect to x can be written as:
d/dx (xy) = x(dy/dx) + y
The derivative of ln(x cot y) with respect to x can be calculated using the chain rule. Let u = x cot y, then:
d/dx (ln x cot y) = d/dx (ln u)
= (1/u) * du/dx
Now, let's calculate du/dx by differentiating both sides of the equation u = x cot y:
du/dx = cot y - x csc^2 y * dy/dx
Substituting the value of du/dx and u into the equation d/dx (ln x cot y), we have:
(1/u) * du/dx = (1/(x cot y)) * (cot y - x csc^2 y * dy/dx)
= (cot y - x csc^2 y * dy/dx) / (x cot y)
Now, equate this expression to the derivative of xy obtained earlier:
x(dy/dx) + y = (cot y - x csc^2 y * dy/dx) / (x cot y)
Multiplying both sides by (x cot y), we get:
x(dy/dx)(x cot y) + y(x cot y) = cot y - x csc^2 y * dy/dx
Expanding and simplifying the equation:
x^2 cot^2 y (dy/dx) + xy cot y + xy = cot y - x csc^2 y * dy/dx
Rearranging the terms:
x^2 cot^2 y (dy/dx) + x csc^2 y * dy/dx = cot y - xy cot y - xy
Factoring out dy/dx:
(dy/dx)(x^2 cot^2 y + x csc^2 y) = cot y - xy cot y - xy
Dividing both sides by (x^2 cot^2 y + x csc^2 y):
dy/dx = (cot y - xy cot y - xy) / (x^2 cot^2 y + x csc^2 y)
Rewriting the right-hand side to get the given expression:
dy/dx = (1-xy) cot y / (x^2 cot^2 y + x csc^2 y)
Hence, we have shown that dy/dx is equal to [(1-xy) cot y] / [x^2 cot^2 y + x csc^2 y] using implicit differentiation.