Analyzing the Setup
Imagine you are standing before a complex, winding path on a graph defined by the equation 2xy+3yx=20. This is an implicit curve where x and y are locked in a dance, appearing in both the base and the exponent.
Our mission is to find the slope of the tangent line to this curve at the specific point (2,2). In the language of calculus, we are hunting for the value of dxdy at that coordinate.
The Power of the uv Formula
When you see a variable raised to a variable power, like xy, the standard power rule dxd(xn)=nxn−1 fails because the exponent is not a constant. We need a more robust tool.
We use the derivative formula for uv, where u and v are both functions of x:
dxd(uv)=uv(uvdxdu+lnudxdv)
This formula is our secret weapon. It allows us to differentiate terms like xy and yx without the headache of taking logs of the entire equation.
Differentiating Term by Term
Let us apply this to our first term, 2xy. Keeping the constant 2 aside, we identify u=x and v=y. The derivative becomes:
2⋅xy(xy⋅1+lnx⋅y′)=2xy(xy+y′lnx)
Now, we turn our attention to the second term, 3yx. Here, the base is y and the exponent is x. Applying the same logic, we get:
3⋅yx(yx⋅y′+lny⋅1)=3yx(y′yx+lny)
Adding these together and setting the result to the derivative of the constant 20 (which is 0), we obtain the full differentiated equation:
2xy(xy+y′lnx)+3yx(y′yx+lny)=0
The Efficiency Shortcut
Instead of expanding this massive expression, we substitute x=2 and y=2 immediately. This collapses the variables into numbers.
Substituting these values, we get:
2(22)(22+y′ln2)+3(22)(y′22+ln2)=0
This simplifies beautifully to:
Expanding this, we get 8+8y′ln2+12y′+12ln2=0.
The Final Stretch
We group the terms containing y′ on one side:
Solving for y′, we get:
Factoring out a 4 from the numerator and denominator, we find:
Finally, applying the power rule of logarithms, nlna=lnan, we transform 3ln2 into ln8 and 2ln2 into ln4. Our final result is: