What is the derivative of f(x)=ln(sqrt(x+9)) ?

The derivative of f(x)=ln(sqrt(x+9)) is 1/(2(x+9))

What is the first derivative of f(x)=ln(sqrt(x+9)) ?

The first derivative of f(x)=ln(sqrt(x+9)) is 1/(2(x+9))

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derivative of

f (x) = ln (x + 9)

\frac{1}{2 ( x + 9 )}

Solution steps

\frac{d}{d x} (ln (x + 9))

Apply the chain rule:

\frac{1}{x + 9} \frac{d}{d x} (x + 9)

= \frac{1}{x + 9} \frac{d}{d x} (x + 9)

\frac{d}{d x} (x + 9) = \frac{1}{2 x + 9}

= \frac{1}{x + 9} \cdot \frac{1}{2 x + 9}

Simplify

\frac{1}{x + 9} \cdot \frac{1}{2 x + 9} : \frac{1}{2 ( x + 9 )}

= \frac{1}{2 ( x + 9 )}

- 4 ln (+ \frac{1}{x} - 4)

(x^{2} + 2 y^{2}) d x = x y d y

\int \frac{sin ( \frac{1}{x ^{4}} )}{x ^{5}} d x

\int_{0.2}^{0.4} 150 x d x

\frac{5}{x + 5}

What is the derivative of f(x)=ln(sqrt(x+9)) ?
The derivative of f(x)=ln(sqrt(x+9)) is 1/(2(x+9))
What is the first derivative of f(x)=ln(sqrt(x+9)) ?
The first derivative of f(x)=ln(sqrt(x+9)) is 1/(2(x+9))