Question

Y=(x+1)^2÷x diffrenciate?

Answer 1

Answer:

2*(x+1)/x - (x+1)^2/x^2

Explanation:

x*y = (x+1)^2 now differentiating both sides

by d/dx

y*dx/dx + x*dy/dx = 2*(x+1)

y + x*dy/dx = 2*(x+1)

putting back value of y,

x*dy/dx = 2*(x+1) - (x+1)^2/x

dy/dx = 2*(x+1)/x - (x+1)^2/x^2

Answer 2

Given:

$\bf{Y=\dfrac{(x+1)^{2}}{x}}$

To Find:

Differentiation of y w.r.t x or $\bf{\dfrac{dy}{dx}}$

Solution:

We know that,

• $\bf{\dfrac{d}{dx}\bigg(\dfrac{f(x)}{g(x)}\bigg)=\dfrac{g(x).f'(x)-f(x)g'(x)}{(g(x))^{2}}}$

where,

f'(x) is $\mathrm{\dfrac{d}{dx}(f(x))}$

g'(x) is $\mathrm{\dfrac{d}{dx}(g(x))}$

• $\bf{\dfrac{d}{dx}(x^{n})=nx^{n-1}}$

Now, on differentiating Y with respect to x, we get

$\longrightarrow\mathrm{\dfrac{dY}{dx}=\dfrac{x.2(x+1)-(x+1)^{2}.1}{x^{2}}}$

$\longrightarrow\mathrm{\dfrac{dY}{dx}=\dfrac{2x(x+1)-(x+1)^{2}}{x^{2}}}$

$\longrightarrow\mathrm{\dfrac{dY}{dx}=\dfrac{2x^{2}+2x-(x^{2}+1+2x)}{x^{2}}}$

$\longrightarrow\mathrm{\dfrac{dY}{dx}=\dfrac{2x^{2}+2x-x^{2}-1-2x}{x^{2}}}$

$\longrightarrow\mathrm{\pink{\dfrac{dY}{dx}=\dfrac{x^{2}-1}{x^{2}}}}$

Hence, differentiation of Y is $\bf{\green{\dfrac{x^{2}-1}{x^{2}}}}$.