Web30 mrt. 2024 Β· Question 16 (OR 1st question) If y = π₯^sinπ₯ +sinγ (π₯^π₯)γ, find ππ¦/ππ₯ Let u = π₯^sinπ₯ , π£=sinγ (π₯^π₯)γ Thus, y = u + v Differentiating π€.π.π‘.π₯ ππ¦/ππ₯ = ππ’/ππ₯ + ππ£/ππ₯ Calculating derivative of u and v separately Solving π π/π π u = π₯^sinπ₯ Taking log both sides logπ’ = log π₯^sinπ₯ logπ’ = sinπ₯ . log π₯ Differentiating π€.π.π‘.π₯ (π (logγπ’)γ)/ππ₯ = π/ππ₯ (sinγπ₯ β¦ Web25 mei 2024 Β· Step-by-step explanation: Given , y = x logx To differentiate or finding dy/dx. We know that, d/dx (uv) = u dv/dx + v du/dx Therefore, we will get, => dy/dx = x d/dx (logx) + logx d/dx (x) But, we know that, d/dx (logx) = 1/x d/dx (x) = 1 Therefore, we will get, => dy/dx = x (1/x) + logx (1) => dy/dx = 1 + logx
Find `Dy/Dx` of Function Xy + Yx = 1 - Mathematics
Webx x (1 + log β‘ x) Finding the value of dy dx: The given function is y = x x. Taking log on both sides, log y = x log x. Differentiating with respect to x. 1 y dy dx = log x + x Γ 1 x [β΅ β¦ Web30 mrt. 2024 Β· Transcript. Ex 5.5, 15 Find ππ¦/ππ₯ of the functions in, π₯π¦= π^ ( (π₯ βπ¦))Given π₯π¦= π^ ( (π₯ βπ¦)) Taking log both sides log (π₯π¦) = log π^ ( (π₯ βπ¦)) log (π₯π¦) = (π₯ βπ¦) log π log π₯+logβ‘π¦ = (π₯ βπ¦) (1) log π₯+logβ‘π¦ β¦ mowry smith iii
1) Find dy dx 2x+11 = (2) If y=ax+b then dxdy β (3) β¦
Web9 jun. 2024 Β· If x y + y x = a b, find dy/dx. cbse; class-12; Share It On Facebook Twitter Email. 1 Answer +1 vote . answered Jun 9, 2024 by paayal (148k points) selected Jun β¦ Web20 apr. 2024 Β· Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto β¦ Weby = x x.. Taking logarithm of both sides, we get. log y = log (x x). β΄ log y = x log x. Differentiating both sides w.r.t.x, we get `1/"y" * "dy"/"dx" = "x" * "d ... mowrys quick fix