What is the derivative of f(x)= x/(sqrt(x^2+7)) ?

The derivative of f(x)= x/(sqrt(x^2+7)) is 7/((x^2+7)sqrt(x^2+7))

What is the first derivative of f(x)= x/(sqrt(x^2+7)) ?

The first derivative of f(x)= x/(sqrt(x^2+7)) is 7/((x^2+7)sqrt(x^2+7))

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derivative of

f (x) = \frac{x}{x ^{2} + 7}

\frac{7}{( x ^{2} + 7 ) x ^{2} + 7}

Solution steps

\frac{d}{d x} (\frac{x}{x ^{2} + 7})

Apply the Quotient Rule:

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

= \frac{\frac{d x}{d x} x ^{2} + 7 - \frac{d}{d x} ( x ^{2} + 7 ) x}{( x ^{2} + 7 ) ^{2}}

\frac{d x}{d x} = 1

\frac{d}{d x} (x^{2} + 7) = \frac{x}{x ^{2} + 7}

= \frac{1 \cdot x ^{2} + 7 - \frac{x}{x ^{2} + 7} x}{( x ^{2} + 7 ) ^{2}}

Simplify

\frac{1 \cdot x ^{2} + 7 - \frac{x}{x ^{2} + 7} x}{( x ^{2} + 7 ) ^{2}} : \frac{7}{( x ^{2} + 7 ) x ^{2} + 7}

= \frac{7}{( x ^{2} + 7 ) x ^{2} + 7}

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

y^{^{''}} + 6 y^{^{'}} + 9 y = 1 + x

\frac{d}{d x} (I n x^{3})

(4, - 9), (8, - 3)

\frac{d ^{2} x}{d t ^{2}} - 4 \frac{d x}{d t} = 4 x

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