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Popular >

derivative of y=ln(x+sqrt(x^2+9))

Solution

derivative of

y = ln (x + x^{2} + 9)

Solution

\frac{1}{x ^{2} + 9}

Solution steps

\frac{d}{d x} (ln (x + x^{2} + 9))

Apply the chain rule:

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

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

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

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

Simplify

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

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

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n = 1 \sum \infty 2^{\frac{1}{n}}

(\partial)/(\partial x)(3x^2+2xy)

\frac{\partial}{\partial x} (3 x^{2} + 2 x y)

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\int \frac{1}{4 ^{x}} d x

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\int x e^{- 6 x^{2}} d x

Frequently Asked Questions (FAQ)

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