What is the derivative of 16-sqrt(x^2+64) ?

The derivative of 16-sqrt(x^2+64) is -x/(sqrt(x^2+64))

What is the first derivative of 16-sqrt(x^2+64) ?

The first derivative of 16-sqrt(x^2+64) is -x/(sqrt(x^2+64))

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\frac{d}{d x} (16 - x^{2} + 64)

- \frac{x}{x ^{2} + 64}

Solution steps

\frac{d}{d x} (16 - x^{2} + 64)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (16) - \frac{d}{d x} (x^{2} + 64)

\frac{d}{d x} (16) = 0

\frac{d}{d x} (x^{2} + 64) = \frac{x}{x ^{2} + 64}

= 0 - \frac{x}{x ^{2} + 64}

Simplify

= - \frac{x}{x ^{2} + 64}

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

\int_{0}^{1} \frac{( x - 1 )}{( x + 1 )} d x

\int 9 x^{2} e^{x} d x

x^{2} = e^{x - y}

\frac{d}{d x} (cos (x^{7}))

What is the derivative of 16-sqrt(x^2+64) ?
The derivative of 16-sqrt(x^2+64) is -x/(sqrt(x^2+64))
What is the first derivative of 16-sqrt(x^2+64) ?
The first derivative of 16-sqrt(x^2+64) is -x/(sqrt(x^2+64))