What is the derivative of (16x)/(x^2+16) ?

The derivative of (16x)/(x^2+16) is (16(-x^2+16))/((x^2+16)^2)

What is the first derivative of (16x)/(x^2+16) ?

The first derivative of (16x)/(x^2+16) is (16(-x^2+16))/((x^2+16)^2)

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derivative of

\frac{16 x}{x ^{2} + 16}

\frac{16 ( - x ^{2} + 16 )}{( x ^{2} + 16 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{16 x}{x ^{2} + 16})

Take the constant out:

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

= 16 \frac{d}{d x} (\frac{x}{x ^{2} + 16})

Apply the Quotient Rule:

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

= 16 \cdot \frac{\frac{d x}{d x} ( x ^{2} + 16 ) - \frac{d}{d x} ( x ^{2} + 16 ) x}{( x ^{2} + 16 ) ^{2}}

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

\frac{d}{d x} (x^{2} + 16) = 2 x

= 16 \cdot \frac{1 \cdot ( x ^{2} + 16 ) - 2 xx}{( x ^{2} + 16 ) ^{2}}

Simplify

16 \cdot \frac{1 \cdot ( x ^{2} + 16 ) - 2 xx}{( x ^{2} + 16 ) ^{2}} : \frac{16 ( - x ^{2} + 16 )}{( x ^{2} + 16 ) ^{2}}

= \frac{16 ( - x ^{2} + 16 )}{( x ^{2} + 16 ) ^{2}}

7 x^{2} e^{x}

f (x) = - 4 (2 x + 7)^{- 5} (2), at x = 2

\int \frac{1}{x 4 x + 9} d x

cos^{5} (x)

x \to 0 lim (\frac{x - 1}{3 x - 1})

What is the derivative of (16x)/(x^2+16) ?
The derivative of (16x)/(x^2+16) is (16(-x^2+16))/((x^2+16)^2)
What is the first derivative of (16x)/(x^2+16) ?
The first derivative of (16x)/(x^2+16) is (16(-x^2+16))/((x^2+16)^2)