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

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

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

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

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derivative of

y = 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 )}

\frac{d}{d x} (4 x^{3} y)

x, 5 x - x^{2}, 0, 4

\int_{- 1}^{1} y^{2} - 1 - (1 - y^{2}) d y

y^{^{'}} = - 2 x y^{3}, y (- 3) = 4

x \to - 6 - lim (\frac{( x + 3 )}{x + 6})

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