Question

y= Ln x + e^x first and second derivative
. plz answer​

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

First derivative,  $\frac{dy}{dx} = \frac{1}{x} + e^{x}$

Second derivative,  $\frac{d^{2}y }{dx^{2} } = \frac{-1}{x^{2} } + e^{x}$

Explanation:

Given:

$y = \ln x + e^{x}$

Find:

First derivative,  $\frac{dy}{dx} = f'(x) = \ ?$

Second derivative,  $\frac{d^{2}y}{dx^{2} } = f''(x) = \ ?$

Solution:

$y = \ln x + e^{x}$

Derivating with respect to $x$, i.e., first derivative,

⇒  $\frac{dy}{dx} = \frac{1}{x} + e^{x}$

Again derivating with respect to $x$, i.e., second derivative,

⇒  $\frac{d^{2}y }{dx^{2} } = \frac{-1}{x^{2} } + e^{x}$

Note:

Derivates for the functions given:

• $\frac{d}{dx} (\ln x) = \frac{1}{x}$
• $\frac{d}{dx} (e^{x} ) = e^{x}$
• $\frac{d}{dx} (\frac{1}{x} ) = \frac{-1}{x^{2} }$

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