Question

find dy/dx when x^2+y^2=sinxy

Answer 1

Answer:

$\displaystyle{\boxed{\red{\sf\:\dfrac{dy}{dx}\:=\:\dfrac{\cos\:(\:xy\:)\:y\:-\:2x}{2y\:-\:x\:\cos\:(\:xy\:)}}}}$

Step-by-step-explanation:

The given function is

$\displaystyle{\sf\:x^2\:+\:y^2\:=\:\sin\:(\:xy\:)}$

We have to find $\displaystyle{\sf\:\dfrac{dy}{dx}}$

Now,

$\displaystyle{\sf\:x^2\:+\:y^2\:=\:\sin\:(\:xy\:)}$

Differentiating both sides w.r.t x, we get,

$\displaystyle{\implies\sf\:\dfrac{d}{dx}\:(\:x^2\:+\:y^2\:)\:=\:\dfrac{d}{dx}\:[\:\sin\:(\:xy\:)\:]}$

We know that,

$\displaystyle{\boxed{\pink{\sf\:\dfrac{d}{dx}\:(\:u\:+\:v\:)\:=\:\dfrac{d}{dx}\:(\:u\:)\:+\:\dfrac{d}{dx}\:(\:v\:)}}}$

$\displaystyle{\implies\sf\:\dfrac{d}{dx}\:(\:x^2\:)\:+\:\dfrac{d}{dx}\:(\:y^2\:)\:=\:\dfrac{d}{dx}\:[\:\sin\:(\:xy\:)\:]}$

We know that,

$\displaystyle{\boxed{\blue{\sf\:\dfrac{d}{dx}\:(\:x^n\:)\:=\:n\:x^{n\:-\:1}}}}$

$\displaystyle{\implies\sf\:2x\:+\:\dfrac{d}{dx}\:(\:y^2\:)\:=\:\dfrac{d}{dx}\:[\:\sin\:(\:xy\:)\:]}$

We know that,

$\displaystyle{\boxed{\green{\sf\:\dfrac{du}{dv}\:=\:\dfrac{du}{dx}\:\times\:\dfrac{dx}{dv}}}}$

$\displaystyle{\implies\sf\:2x\:+\:\dfrac{d}{dy}\:(\:y^2\:)\:\times\:\dfrac{dy}{dx}\:=\:\dfrac{d}{dx}\:[\:\sin\:(\:xy\:)\:]}$

$\displaystyle{\implies\sf\:2x\:+\:2y\:\dfrac{dy}{dx}\:=\:\dfrac{d}{dx}\:[\:\sin\:(\:xy\:)\:]}$

We know that,

$\displaystyle{\boxed{\orange{\sf\:\dfrac{d}{dx}\:(\:\sin\:x\:)\:=\:\cos\:(\:x\:)}}}$

$\displaystyle{\boxed{\purple{\sf\:\dfrac{d}{dx}\:[\:f\:(\:g\:(\:x\:)\:)\:]\:=\:\dfrac{d}{dx}\:[\:f\:(\:x\:)\:]\:.\:\dfrac{d}{dx}\:[\:g\:(\:x\:)\:]}}}$

$\displaystyle{\implies\sf\:2x\:+\:2y\:\dfrac{dy}{dx}\:=\:\cos\:(\:xy\:)\:.\:\dfrac{d}{dx}\:(\:xy\:)}$

We know that,

$\displaystyle{\boxed{\pink{\sf\:\dfrac{d}{dx}\:(\:uv\:)\:=\:v\:\dfrac{d}{dx}\:(\:u\:)\:+\:u\:\dfrac{d}{dx}\:(\:v\:)}}}$

$\displaystyle{\implies\sf\:2x\:+\:2y\:\dfrac{dy}{dx}\:=\:\cos\:(\:xy\:)\:\left[\:y\:\dfrac{d}{dx}\:(\:x\:)\:+\:x\:\dfrac{d}{dx}\:(\:y\:)\:\right]}$

$\displaystyle{\implies\sf\:2x\:+\:2y\:\dfrac{dy}{dx}\:=\:\cos\:(\:xy\:)\:\left[\:y\:\times\:1\:+\:x\:\dfrac{d}{dy}\:(\:y\:)\:\times\:\dfrac{dy}{dx}\:\right]}$

$\displaystyle{\implies\sf\:2x\:+\:2y\:\dfrac{dy}{dx}\:=\:\cos\:(\:xy\:)\:\left[\:y\:+\:x\:\times\:1\:\dfrac{dy}{dx}\:\right]}$

$\displaystyle{\implies\sf\:2x\:+\:2y\:\dfrac{dy}{dx}\:=\:\cos\:(\:xy\:)\:\left[\:y\:+\:x\:\dfrac{dy}{dx}\:\right]}$

$\displaystyle{\implies\sf\:2x\:+\:2y\:\dfrac{dy}{dx}\:=\:\cos\:(\:xy\:)\:y\:+\:x\:\dfrac{dy}{dx}\:\cos\:(\:xy\:)}$

$\displaystyle{\implies\sf\:2y\:\dfrac{dy}{dx}\:-\:x\:\cos\:(\:xy\:)\:\dfrac{dy}{dx}\:=\:\cos\:(\:xy\:)\:y\:-\:2x}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:[\:2y\:-\:x\:\cos\:(\:xy\:)\:]\:=\:\cos\:(\:xy\:)\:y\:-\:2x}$

$\displaystyle{\implies\:\underline{\boxed{\red{\sf\:\dfrac{dy}{dx}\:=\:\dfrac{\cos\:(\:xy\:)\:y\:-\:2x}{2y\:-\:x\:\cos\:(\:xy\:)}}}}}$

Answer 2

★Question

➽ Find dy/dx when x²+y² = sin (xy)

★ Solution

★ Given

➽ x²+y²= sin (xy)

To find

➽ dy/dx

So,

➽ x²+y² = sin (xy)

We know that

➽ d/dx (u+v) = d/dx (u) + d/dx (v)

➽ d/dx (x²)+d/dx(y²) = d/dx [ sin (xy) ]

★ Now,

➽ d/dx (xⁿ) = n^-1

➽ 2x+d/dx(y²) = d/dx [sin(xy)]

★We know that

➽ du/dv= du/dx × dx/dv

➽ 2x + d/dy(y²) × dy/dx = d/dx [sin (xy)]

➽ 2x + 2y dy/dx = d/dx [ sin (xy)]

Now,

➽ d/dx(sin x) = cos (x)

➽ dx/dx[f(g(x))]= d/dz[f(x)]•d/dx[g(x)]

➽ 2x+2y dy/dx = cos (xy)/d/dx (xy)

We know that,

➽ d/dz(uv) = v d/dx (u) + u d/dx (v)

➠ 2x+2y dy/dx = cos (xy) [ y d/dx (x) +x d/dy (y) ]

➠ 2x+2dy/dx = cos (xy)[y×1+x d/dy (y) × dy/dx]

➠ 2x+2y dy/dx = cos (xy) [ y+x dy/dx ]

➠ 2x+2y dy/dx = cos (xy ) [ y+x dy/dx]

➠ 2x+2y dy/dx = cos (xy)y + x dy/dx cos (xy)

➠ 2y dy/dx - x cos (xy) dy/dx = cos (xy) y-2x

➠ dy/dz[ 2y -x cos (xy) ] = cos (xy) y-2x

$\to \tt \: answer = \green{ \frac{ cos \: (xy)y - 2x}{2y - x \: cos \: (xy)} }$