What is the derivative of f(x)= x/(2(x^2+7)^{3/4)} ?

The derivative of f(x)= x/(2(x^2+7)^{3/4)} is (-x^2+14)/(4(x^2+7)^{7/4)}

What is the first derivative of f(x)= x/(2(x^2+7)^{3/4)} ?

The first derivative of f(x)= x/(2(x^2+7)^{3/4)} is (-x^2+14)/(4(x^2+7)^{7/4)}

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derivative of

f (x) = \frac{x}{2 ( x ^{2} + 7 ) ^{\frac{3}{4}}}

\frac{- x ^{2} + 14}{4 ( x ^{2} + 7 ) ^{\frac{7}{4}}}

Solution steps

\frac{d}{d x} (\frac{x}{2 ( x ^{2} + 7 ) ^{\frac{3}{4}}})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{2} \frac{d}{d x} (\frac{x}{( x ^{2} + 7 ) ^{\frac{3}{4}}})

Apply the Quotient Rule:

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

= \frac{1}{2} \cdot \frac{\frac{d x}{d x} ( x ^{2} + 7 ) ^{\frac{3}{4}} - \frac{d}{d x} ( ( x ^{2} + 7 ) ^{\frac{3}{4}} ) x}{( ( x ^{2} + 7 ) ^{\frac{3}{4}} ) ^{2}}

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

\frac{d}{d x} ((x^{2} + 7)^{\frac{3}{4}}) = \frac{3 x}{2 ( x ^{2} + 7 ) ^{\frac{1}{4}}}

= \frac{1}{2} \cdot \frac{1 \cdot ( x ^{2} + 7 ) ^{\frac{3}{4}} - \frac{3 x}{2 ( x ^{2} + 7 ) ^{\frac{1}{4}}} x}{( ( x ^{2} + 7 ) ^{\frac{3}{4}} ) ^{2}}

Simplify

\frac{1}{2} \cdot \frac{1 \cdot ( x ^{2} + 7 ) ^{\frac{3}{4}} - \frac{3 x}{2 ( x ^{2} + 7 ) ^{\frac{1}{4}}} x}{( ( x ^{2} + 7 ) ^{\frac{3}{4}} ) ^{2}} : \frac{- x ^{2} + 14}{4 ( x ^{2} + 7 ) ^{\frac{7}{4}}}

= \frac{- x ^{2} + 14}{4 ( x ^{2} + 7 ) ^{\frac{7}{4}}}

\frac{d}{d x} (sin (5 x + 2))

f (x) = \frac{x + 1}{x}

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

f (x) = e^{9}

What is the derivative of f(x)= x/(2(x^2+7)^{3/4)} ?
The derivative of f(x)= x/(2(x^2+7)^{3/4)} is (-x^2+14)/(4(x^2+7)^{7/4)}
What is the first derivative of f(x)= x/(2(x^2+7)^{3/4)} ?
The first derivative of f(x)= x/(2(x^2+7)^{3/4)} is (-x^2+14)/(4(x^2+7)^{7/4)}