What is the derivative of y= 1/(2sqrt(x+7)) ?

The derivative of y= 1/(2sqrt(x+7)) is -1/(4(x+7)^{3/2)}

What is the first derivative of y= 1/(2sqrt(x+7)) ?

The first derivative of y= 1/(2sqrt(x+7)) is -1/(4(x+7)^{3/2)}

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derivative of

y = \frac{1}{2 x + 7}

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

Solution steps

\frac{d}{d x} (\frac{1}{2 x + 7})

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

\frac{d}{d x} (x + 7) = 1

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

Simplify

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

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

f (x) = 2 x + \frac{5}{x}

ln (2 - x^{2})

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

\int \frac{y}{x ^{2}} d x

x \to 9 lim (x - 3)

What is the derivative of y= 1/(2sqrt(x+7)) ?
The derivative of y= 1/(2sqrt(x+7)) is -1/(4(x+7)^{3/2)}
What is the first derivative of y= 1/(2sqrt(x+7)) ?
The first derivative of y= 1/(2sqrt(x+7)) is -1/(4(x+7)^{3/2)}