What is the derivative of ((x+4^2)/(x+3)) ?

The derivative of ((x+4^2)/(x+3)) is (x^2+6x+8)/((x+3)^2)

What is the first derivative of ((x+4^2)/(x+3)) ?

The first derivative of ((x+4^2)/(x+3)) is (x^2+6x+8)/((x+3)^2)

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\frac{d}{d x} (\frac{( x + 4 ) ^{2}}{x + 3})

\frac{x ^{2} + 6 x + 8}{( x + 3 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{( x + 4 ) ^{2}}{x + 3})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( ( x + 4 ) ^{2} ) ( x + 3 ) - \frac{d}{d x} ( x + 3 ) ( x + 4 ) ^{2}}{( x + 3 ) ^{2}}

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

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

= \frac{2 ( x + 4 ) ( x + 3 ) - 1 \cdot ( x + 4 ) ^{2}}{( x + 3 ) ^{2}}

Simplify

\frac{2 ( x + 4 ) ( x + 3 ) - 1 \cdot ( x + 4 ) ^{2}}{( x + 3 ) ^{2}} : \frac{x ^{2} + 6 x + 8}{( x + 3 ) ^{2}}

= \frac{x ^{2} + 6 x + 8}{( x + 3 ) ^{2}}

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

y d x - (y - x) d y = 0

y^{^{'}} + \frac{1}{2} y = 0

y^{^{''}} + 10 y^{^{'}} + y = 0

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

What is the derivative of ((x+4^2)/(x+3)) ?
The derivative of ((x+4^2)/(x+3)) is (x^2+6x+8)/((x+3)^2)
What is the first derivative of ((x+4^2)/(x+3)) ?
The first derivative of ((x+4^2)/(x+3)) is (x^2+6x+8)/((x+3)^2)