How do you implicitly differentiate #-1=x-ycot^2(x-y) #?

Question

How do you implicitly differentiate #-1=x-ycot^2(x-y)  #?

Samantha Adkison · Accepted Answer

#(dy)/dx =(1+2ycot(x - y) csc^2(x - y))/(cot^2(x-y)+2ycot(x - y) csc^2(x - y))# Take the derivative, #d/dx# of both sides #d/dx (-1)=d/dx (x)-d/dx(ycot^2(x-y))# #0=1-yd/dx(cot^2(x-y))-(dy)/dx cot^2(x-y)# #(dy)/dx cot^2(x-y)+yd/dx(cot^2(x-y))=1# #(dy)/dx cot^2(x-y)+y(-2 cot(x - y) csc^2(x - y))(1-dy/dx)=1# #(dy)/dx cot^2(x-y)-2ycot(x - y) csc^2(x - y)+2ycot(x - y) csc^2(x - y)dy/dx=1# #(dy)/dx cot^2(x-y)+dy/dx2ycot(x - y) csc^2(x - y)-2ycot(x - y) csc^2(x - y)=1# #(dy)/dx (cot^2(x-y)+2ycot(x - y) csc^2(x - y))-2ycot(x - y) csc^2(x - y)=1# #(dy)/dx (cot^2(x-y)+2ycot(x - y) csc^2(x - y))=1+2ycot(x - y) csc^2(x - y)# #(dy)/dx =(1+2ycot(x - y) csc^2(x - y))/(cot^2(x-y)+2ycot(x - y) csc^2(x - y))#

Charles Anctil · Answer

undefined To implicitly differentiate the equation $-1 = x - y \cot^2(x - y)$, follow these steps:

1. Differentiate both sides of the equation with respect to $x$.
2. Apply the chain rule and product rule as necessary.
3. Solve for $\frac{{dy}}{{dx}}$.

Differentiating $-1 = x - y \cot^2(x - y)$ with respect to $x$ yields:

$$0 = 1 - \frac{{dy}}{{dx}} \cot^2(x - y) - y \cdot \frac{{d}}{{dx}}[\cot^2(x - y)]$$

Using the chain rule, differentiate $\cot^2(x - y)$ with respect to $x$:

$$\frac{{d}}{{dx}}[\cot^2(x - y)] = -2\cot(x - y)\csc^2(x - y) \cdot \frac{{d}}{{dx}}(x - y)$$

Simplify:

$$\frac{{d}}{{dx}}(x - y) = 1 - \frac{{dy}}{{dx}}$$

Substitute the expression for $\frac{{d}}{{dx}}(x - y)$ into the equation:

$$0 = 1 - \frac{{dy}}{{dx}} \cot^2(x - y) - y \cdot (-2\cot(x - y)\csc^2(x - y))$$

Now, solve for $\frac{{dy}}{{dx}}$.