What is the derivative of f(x)=x^2(3+x)^{-5} ?

The derivative of f(x)=x^2(3+x)^{-5} is (-3x^2+6x)/((x+3)^6)

What is the first derivative of f(x)=x^2(3+x)^{-5} ?

The first derivative of f(x)=x^2(3+x)^{-5} is (-3x^2+6x)/((x+3)^6)

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derivative of

f (x) = x^{2} (3 + x)^{- 5}

\frac{- 3 x ^{2} + 6 x}{( x + 3 ) ^{6}}

Solution steps

\frac{d}{d x} (x^{2} (3 + x)^{- 5})

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{2}, g = (3 + x)^{- 5}

= \frac{d}{d x} (x^{2}) (3 + x)^{- 5} + \frac{d}{d x} ((3 + x)^{- 5}) x^{2}

\frac{d}{d x} (x^{2}) = 2 x

\frac{d}{d x} ((3 + x)^{- 5}) = - \frac{5}{( 3 + x ) ^{6}}

= 2 x (3 + x)^{- 5} + (- \frac{5}{( 3 + x ) ^{6}}) x^{2}

Simplify

2 x (3 + x)^{- 5} + (- \frac{5}{( 3 + x ) ^{6}}) x^{2} : \frac{- 3 x ^{2} + 6 x}{( x + 3 ) ^{6}}

= \frac{- 3 x ^{2} + 6 x}{( x + 3 ) ^{6}}

\frac{d}{d t} (e^{t} (cos (t) - sin (t)))

y^{^{''}} + \frac{1}{4} y = 0, y (π) = 1, y^{^{'}} (π) = - 3

\frac{d}{d x} ((\frac{x - 1}{x ^{2} + x + 1})^{4})

\int sin (x) + 3 d x

t an g e n t f (x) = x^{2} - x, at x = 3

What is the derivative of f(x)=x^2(3+x)^{-5} ?
The derivative of f(x)=x^2(3+x)^{-5} is (-3x^2+6x)/((x+3)^6)
What is the first derivative of f(x)=x^2(3+x)^{-5} ?
The first derivative of f(x)=x^2(3+x)^{-5} is (-3x^2+6x)/((x+3)^6)