Math / Calculus

QuestionIf $y=5 x \sqrt{x^{2}+1}$ , then $\frac{d y}{d x}$ at $x=3$ is

Studdy Solution

STEP 1

1. We are given the function $y = 5x \sqrt{x^2 + 1}$ .
2. We need to find the derivative $\frac{dy}{dx}$ at $x = 3$ .

STEP 2

1. Identify the differentiation technique required.
2. Differentiate the function $y$ with respect to $x$ .
3. Simplify the derivative expression.
4. Substitute $x = 3$ into the derivative.
5. Calculate the value of the derivative at $x = 3$ .

STEP 3

Identify the differentiation technique required. The function $y = 5x \sqrt{x^2 + 1}$ is a product of two functions: $u = 5x$ and $v = \sqrt{x^2 + 1}$ . We will use the product rule for differentiation.

STEP 4

Apply the product rule: $\frac{d}{dx}(uv) = u'v + uv'$ .
First, differentiate $u = 5x$ :
$u' = \frac{d}{dx}(5x) = 5$
Next, differentiate $v = \sqrt{x^2 + 1}$ :
$v = (x^2 + 1)^{1/2}$
Using the chain rule, differentiate $v$ :
$v' = \frac{1}{2}(x^2 + 1)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 + 1}}$

STEP 5

Substitute $u$ , $u'$ , $v$ , and $v'$ into the product rule:
$\frac{dy}{dx} = u'v + uv'$ $\frac{dy}{dx} = 5 \cdot \sqrt{x^2 + 1} + 5x \cdot \frac{x}{\sqrt{x^2 + 1}}$
Simplify the expression:
$\frac{dy}{dx} = 5\sqrt{x^2 + 1} + \frac{5x^2}{\sqrt{x^2 + 1}}$
Combine the terms:
$\frac{dy}{dx} = \frac{5(x^2 + 1) + 5x^2}{\sqrt{x^2 + 1}}$ $\frac{dy}{dx} = \frac{5x^2 + 5 + 5x^2}{\sqrt{x^2 + 1}}$ $\frac{dy}{dx} = \frac{10x^2 + 5}{\sqrt{x^2 + 1}}$

STEP 6

Substitute $x = 3$ into the derivative:
$\frac{dy}{dx} \bigg|_{x=3} = \frac{10(3)^2 + 5}{\sqrt{(3)^2 + 1}}$

STEP 7

Calculate the value:
$\frac{dy}{dx} \bigg|_{x=3} = \frac{10 \cdot 9 + 5}{\sqrt{9 + 1}}$ $\frac{dy}{dx} \bigg|_{x=3} = \frac{90 + 5}{\sqrt{10}}$ $\frac{dy}{dx} \bigg|_{x=3} = \frac{95}{\sqrt{10}}$
Rationalize the denominator:
$\frac{dy}{dx} \bigg|_{x=3} = \frac{95}{\sqrt{10}} \cdot \frac{\sqrt{10}}{\sqrt{10}}$ $\frac{dy}{dx} \bigg|_{x=3} = \frac{95\sqrt{10}}{10}$
The value of $\frac{dy}{dx}$ at $x = 3$ is:
$\boxed{\frac{95\sqrt{10}}{10}}$

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