JEE Main 2023 · 6 April · morning · Single correct · +4 / −1

If 2 xʸ+3 yˣ=20, then (d y)/(d x) at (2,2) is equal to :

  1. (a) -((3+log ₑ 16)/(4+log ₑ 8))
  2. (b) -((2+log ₑ 8)/(3+log ₑ 4))
  3. (c) -((3+log ₑ 8)/(2+log ₑ 4))
  4. (d) -((3+log ₑ 4)/(2+log ₑ 8))
Answer and solution

Answer (b) -((2+log ₑ 8)/(3+log ₑ 4))

Solution

Maths sir · 4 steps

Write each variable power as an exponential, differentiate, then put in (2,2) before you do any algebra.

  1. Handle variable powers. Write xʸ = eʸˡⁿˣ and yˣ = eˣˡⁿʸ. Differentiating the exponent gives d/dxxʸ = xʸ(y'ln x + y/x) and d/dxyˣ = yˣ(ln y + x/yy').

    2xʸ(y'ln x + y/x) + 3yˣ(ln y + x/y y') = 0

  2. Put in (2,2). Here xʸ = yˣ = 2² = 4 and y/x = x/y = 1. The equation becomes 8(y'ln 2 + 1) + 12(ln 2 + y') = 0.
  3. Solve for y′. Collect the y' terms: y'(12 + 8 ln 2) = -(8 + 12 ln 2). Dividing top and bottom by 4 gives y' = -(2 + 3 ln 2)/(3 + 2 ln 2).
  4. So. Since 3 ln 2 = logₑ 8 and 2 ln 2 = logₑ 4, dy/dx = -((2+logₑ 8)/(3+logₑ 4)), which is option (b).

The idea to keep. When both the base and the power are variables, write uᵛ = eᵛˡⁿᵘ and differentiate the exponent. Put in the point before you solve for y', because that keeps the algebra short.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Differentiation

Asked about 40 times in JEE Main since 2020. This question is from 2023.