Question

Find dy/dx, if (sin x)^y=x+y

If answered correctly, I will mark as brainliest,
If answered incorrectly, I will report

Answer 1

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$\frac{dy}{dx}$ = $\frac{1- y(x+y)^2cot x}{(x+y)[(x+y)log(sinx)-1]}$

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step-by-step explanation:

given,

${(sin x)}^{y}$ = x + y

Now,

taking log on both sides,

we get,

y log(sinx) = log (x+y)

°.° log ${m}^{n}$ = n log m

now,

there are two functions on L.H.S

let,

u = y

and

v = log (sinx)

now,

we know that,

d(uv)/dx = u dv/dx + v du/dx

so,

diffrentiating the given equation on both sides w.r.t x,

we get,

y d{log (sinx)}/dx + log(sinx) dy/dx = d[log(x+y)]/dx ...............(i)

now,

for,

d{log(sinx)}/dx

let,

sinx = t

diffrentiating wrt x on both sides,

we get,

dt/dx = cos x

let,

m = log (sinx) = log t

=> dm/dt = 1/t

.°. dm/ dx = (dm/dt) × (dt/dx)

= 1/t × cosx

= cosx/t

= cosx / sinx

= cot x .....................(ii)

again,

for,

d{log(x+y)}/dx

let,

log(x+y) = a

and,

x +y = b

=> db/dx = 1 + dy/dx

and,

a = log b

=> da/db = 1/b

.°. da/dx = (da/db)× (db/dx)

= 1/b × (1+ dy/dx)

= (1+ dy/dx)/(x+y) .............(iii)

now,

putting the values of differntials from Equation (ii) and (iii) in equation (i),

we get,

y cot x + (dy/dx){log (sin x)} = (1/x+y) + (dy/dx)(1/x+y)

=> (dy/dx){log(sinx) - 1/(x+y)} = 1/(x+y) - ycotx

=> $\frac{dy}{dx}$ = $\frac{1- y(x+y)^2cot x}{(x+y)[(x+y)log(sinx)-1]}$

This question was very tricky which came in CBSE class 12th maths board examination 2019.