What is the derivative of f(-7)= 1/(x+3) ?

The derivative of f(-7)= 1/(x+3) is -1/((x+3)^2)

What is the first derivative of f(-7)= 1/(x+3) ?

The first derivative of f(-7)= 1/(x+3) is -1/((x+3)^2)

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derivative of

f (- 7) = \frac{1}{x + 3}

- \frac{1}{( x + 3 ) ^{2}}

Solution steps

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

- \frac{1}{( x + 3 ) ^{2}} \frac{d}{d x} (x + 3)

= - \frac{1}{( x + 3 ) ^{2}} \frac{d}{d x} (x + 3)

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

= - \frac{1}{( x + 3 ) ^{2}} \cdot 1

Simplify

= - \frac{1}{( x + 3 ) ^{2}}

y = \frac{2}{3 x ^{2}}

x \to 0 lim (\frac{4 x ^{2}}{x})

y^{^{'}} = 9 y^{2} + x y^{2}, y (0) = 1

\frac{\partial}{\partial x} ((41 - x^{2} - y^{2})^{\frac{1}{2}})

\int_{0}^{1} x e^{- 4 x} d x

What is the derivative of f(-7)= 1/(x+3) ?
The derivative of f(-7)= 1/(x+3) is -1/((x+3)^2)
What is the first derivative of f(-7)= 1/(x+3) ?
The first derivative of f(-7)= 1/(x+3) is -1/((x+3)^2)