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Popular >

y^'=(-1)/(x^2)

Solution

y^{^{'}} = \frac{- 1}{x ^{2}}

Solution

y = \frac{1}{x} + c_{1}

Solution steps

y^{'} (x) = \frac{- 1}{x ^{2}}

If

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int \frac{- 1}{x ^{2}} d x

\int \frac{- 1}{x ^{2}} d x = \frac{1}{x} + c_{1}

y = \frac{1}{x} + c_{1}

Popular Examples

sum from n=0 to infinity of 1/5*(-2/3)^n

n = 0 \sum \infty \frac{1}{5} \cdot (- \frac{2}{3})^{n}

tangent of x^2+y^2=25,(-4,3)tangent of

x^{2} + y^{2} = 25, (- 4, 3)

tangent of f(x)=(4x)/(x^2+3),\at x=1 tangent of

f (x) = \frac{4 x}{x ^{2} + 3}, at x = 1

limit as x approaches 0+of cos(pi/x)

x \to 0 + lim (cos (\frac{π}{x}))

derivative of (1+x^{1/3})

\frac{d}{d x} ((1 + x)^{\frac{1}{3}})

Frequently Asked Questions (FAQ)

What is the solution for y^'=(-1)/(x^2) ?
The solution for y^'=(-1)/(x^2) is y= 1/x+c_{1}