Question

x+y=log(x+y) find dy/dx please answer the question step by step​ ,those who don't know please not comment it

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:x + y = log(x + y)$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}(x + y) = \dfrac{d}{dx}log(x + y)$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx}logx = \frac{1}{x} \: }}$

and

$\boxed{ \tt{ \: \dfrac{d}{dx}x = 1 \: }}$

So, using this result, we get

$\rm :\longmapsto\:1 + \dfrac{dy}{dx} = \dfrac{1}{x + y}\dfrac{d}{dx}(x + y)$

$\rm :\longmapsto\:1 + \dfrac{dy}{dx} = \dfrac{1}{x + y}\bigg(1 + \dfrac{dy}{dx} \bigg)$

$\rm :\longmapsto\:1 + \dfrac{dy}{dx} = \dfrac{1}{x + y} + \dfrac{1}{x + y} \dfrac{dy}{dx}$

$\rm :\longmapsto\: \dfrac{dy}{dx} - \dfrac{1}{x + y} \dfrac{dy}{dx} = \dfrac{1}{x + y} - 1$

$\rm :\longmapsto\: \bigg(1 - \dfrac{1}{x + y}\bigg)\dfrac{dy}{dx} = \dfrac{1}{x + y} - 1$

$\rm :\longmapsto\: \bigg(\dfrac{x + y - 1}{x + y}\bigg)\dfrac{dy}{dx} = \dfrac{1 - x - y}{x + y}$

$\rm :\longmapsto\: \bigg(\dfrac{x + y - 1}{x + y}\bigg)\dfrac{dy}{dx} = - \: \dfrac{x + y - 1}{x + y}$

$\rm \implies\:\boxed{ \tt{ \: \dfrac{dy}{dx} \: = \: - \: 1 \: }}$

More to know :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

Answer:

Solution−

Given function is

\rm :\longmapsto\:x + y = log(x + y):⟼x+y=log(x+y)

On differentiating both sides w. r. t. x, we get

\rm :\longmapsto\:\dfrac{d}{dx}(x + y) = \dfrac{d}{dx}log(x + y):⟼

dx

d

(x+y)=

dx

d

log(x+y)

We know,

\boxed{ \tt{ \: \dfrac{d}{dx}logx = \frac{1}{x} \: }}

dx

d

logx=

x

1

and

\boxed{ \tt{ \: \dfrac{d}{dx}x = 1 \: }}

dx

d

x=1

So, using this result, we get

\rm :\longmapsto\:1 + \dfrac{dy}{dx} = \dfrac{1}{x + y}\dfrac{d}{dx}(x + y):⟼1+

dx

dy

=

x+y

1

dx

d

(x+y)

\rm :\longmapsto\:1 + \dfrac{dy}{dx} = \dfrac{1}{x + y}\bigg(1 + \dfrac{dy}{dx} \bigg):⟼1+

dx

dy

=

x+y

1

(1+

dx

dy

)

\rm :\longmapsto\:1 + \dfrac{dy}{dx} = \dfrac{1}{x + y} + \dfrac{1}{x + y} \dfrac{dy}{dx}:⟼1+

dx

dy

=

x+y

1

+

x+y

1

dx

dy

\rm :\longmapsto\: \dfrac{dy}{dx} - \dfrac{1}{x + y} \dfrac{dy}{dx} = \dfrac{1}{x + y} - 1:⟼

dx

dy

−

x+y

1

dx

dy

=

x+y

1

−1

\rm :\longmapsto\: \bigg(1 - \dfrac{1}{x + y}\bigg)\dfrac{dy}{dx} = \dfrac{1}{x + y} - 1:⟼(1−

x+y

1

)

dx

dy

=

x+y

1

−1

\rm :\longmapsto\: \bigg(\dfrac{x + y - 1}{x + y}\bigg)\dfrac{dy}{dx} = \dfrac{1 - x - y}{x + y}:⟼(

x+y

x+y−1

)

dx

dy

=

x+y

1−x−y

\rm :\longmapsto\: \bigg(\dfrac{x + y - 1}{x + y}\bigg)\dfrac{dy}{dx} = - \: \dfrac{x + y - 1}{x + y}:⟼(

x+y

x+y−1

)

dx

dy

=−

x+y

x+y−1

\rm \implies\:\boxed{ \tt{ \: \dfrac{dy}{dx} \: = \: - \: 1 \: }}⟹

dx

dy

=−1

More to know :-

\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}\end{gathered}

f(x)

k

sinx

cosx

tanx

cotx

secx

cosecx

x

logx

e

x

dx

d

f(x)

0

cosx

−sinx

sec

2

x

−cosec

2

x

secxtanx

−cosecxcotx

2

x

1

x

1

e

x