JEE Mathematics 2023 Question with Solution

Given: $2x^y + 3y^x = 20$

Find: $\frac{dy}{dx}$ at $(2,2)$.

Differentiating both sides with respect to $x$:

$$2 \frac{d}{dx}(x^y) + 3 \frac{d}{dx}(y^x) = 0$$

Using the required differentiation formulas:

$$2x^y \left( \frac{y}{x} + \ln(x) \frac{dy}{dx} \right) + 3y^x \left( \frac{x}{y} \frac{dy}{dx} + \ln(y) \right) = 0$$

Substituting $x = 2$ and $y = 2$:

$$8 \left( \frac{1}{2} + \ln(2) \frac{dy}{dx} \right) + 12 \left( \frac{1}{2} \frac{dy}{dx} + \ln(2) \right) = 0$$

Now simplify:

$$4 + 8\ln(2)\frac{dy}{dx} + 6\frac{dy}{dx} + 12\ln(2) = 0$$ $$\left(6 + 8\ln(2)\right)\frac{dy}{dx} = -(4 + 12\ln(2))$$ $$\frac{dy}{dx} = -\frac{4 + 12\ln(2)}{6 + 8\ln(2)} = -\frac{2 + \ln(8)}{3 + \ln(4)}$$

Therefore, $\frac{dy}{dx} = -\frac{2 + \ln(8)}{3 + \ln(4)}$.

Given: $2x^y + 3y^x = 20$

Find: how to differentiate $x^y$ and $y^x$ implicitly.

Use logarithmic differentiation identities:

$$\frac{d}{dx}(x^y) = x^y\left(\frac{y}{x} + \ln(x)\frac{dy}{dx}\right)$$

because

$$\ln(x^y) = y\ln x$$

and differentiating gives

$$\frac{1}{x^y}\frac{d}{dx}(x^y) = \frac{dy}{dx}\ln x + \frac{y}{x}$$

Similarly,

$$\frac{d}{dx}(y^x) = y^x\left(\ln y + \frac{x}{y}\frac{dy}{dx}\right)$$

Substituting these into the differentiated equation and then putting $(x,y)=(2,2)$ leads to

$$\frac{dy}{dx} = -\frac{2 + \ln(8)}{3 + \ln(4)}$$

So the derivative is negative. This means the correct computed value does not match the unsigned option statement from the answer key text, even though the keyed option position is B.