What is the derivative of f(x)=((x-3)^2)/((x+9)) ?

The derivative of f(x)=((x-3)^2)/((x+9)) is (x^2+18x-63)/((x+9)^2)

What is the first derivative of f(x)=((x-3)^2)/((x+9)) ?

The first derivative of f(x)=((x-3)^2)/((x+9)) is (x^2+18x-63)/((x+9)^2)

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derivative of

f (x) = \frac{( x - 3 ) ^{2}}{( x + 9 )}

\frac{x ^{2} + 18 x - 63}{( x + 9 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{( x - 3 ) ^{2}}{x + 9})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( ( x - 3 ) ^{2} ) ( x + 9 ) - \frac{d}{d x} ( x + 9 ) ( x - 3 ) ^{2}}{( x + 9 ) ^{2}}

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

\frac{d}{d x} (x + 9) = 1

= \frac{2 ( x - 3 ) ( x + 9 ) - 1 \cdot ( x - 3 ) ^{2}}{( x + 9 ) ^{2}}

Simplify

\frac{2 ( x - 3 ) ( x + 9 ) - 1 \cdot ( x - 3 ) ^{2}}{( x + 9 ) ^{2}} : \frac{x ^{2} + 18 x - 63}{( x + 9 ) ^{2}}

= \frac{x ^{2} + 18 x - 63}{( x + 9 ) ^{2}}

8 x + 8 x^{- 1}

\frac{d}{d x} (e^{- \frac{x}{y}})

\frac{d}{d x} (\frac{4}{3} p i x^{3})

3 x^{2}, at x = 4

x \to 5 lim (x)

What is the derivative of f(x)=((x-3)^2)/((x+9)) ?
The derivative of f(x)=((x-3)^2)/((x+9)) is (x^2+18x-63)/((x+9)^2)
What is the first derivative of f(x)=((x-3)^2)/((x+9)) ?
The first derivative of f(x)=((x-3)^2)/((x+9)) is (x^2+18x-63)/((x+9)^2)