Math / Calculus

QuestionDifferentiate the function $y=\left(x^{2}+9\right) \ln \left(x^{2}+9\right)$ $y^{\prime}=$

Studdy Solution

STEP 1

1. We are given the function $y = (x^2 + 9) \ln(x^2 + 9)$ .
2. We need to find the derivative $y'$ with respect to $x$ .

STEP 2

1. Identify the structure of the function and determine the differentiation method.
2. Apply the product rule to differentiate the function.
3. Simplify the derivative expression.

STEP 3

Identify the structure of the function. The function $y = (x^2 + 9) \ln(x^2 + 9)$ is a product of two functions: - $u(x) = x^2 + 9$ - $v(x) = \ln(x^2 + 9)$

STEP 4

Apply the product rule for differentiation, which states that if $y = u(x) \cdot v(x)$ , then:
$y' = u'(x) \cdot v(x) + u(x) \cdot v'(x)$

STEP 5

Differentiate $u(x) = x^2 + 9$ :
$u'(x) = \frac{d}{dx}(x^2 + 9) = 2x$

STEP 6

Differentiate $v(x) = \ln(x^2 + 9)$ using the chain rule:
$v'(x) = \frac{d}{dx}[\ln(x^2 + 9)] = \frac{1}{x^2 + 9} \cdot \frac{d}{dx}(x^2 + 9) = \frac{1}{x^2 + 9} \cdot 2x = \frac{2x}{x^2 + 9}$

STEP 7

Substitute $u'(x)$ , $v(x)$ , $u(x)$ , and $v'(x)$ into the product rule formula:
$y' = (2x) \cdot \ln(x^2 + 9) + (x^2 + 9) \cdot \frac{2x}{x^2 + 9}$

STEP 8

Simplify the expression:
$y' = 2x \ln(x^2 + 9) + 2x$
The derivative of the function is:
$y' = 2x \ln(x^2 + 9) + 2x$

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QuestionDifferentiate the function $y=\left(x^{2}+9\right) \ln \left(x^{2}+9\right)$ $y^{\prime}=$

Studdy Solution

STEP 1

1. We are given the function $y = (x^2 + 9) \ln(x^2 + 9)$ .
2. We need to find the derivative $y'$ with respect to $x$ .

STEP 2

1. Identify the structure of the function and determine the differentiation method.
2. Apply the product rule to differentiate the function.
3. Simplify the derivative expression.

STEP 3

Identify the structure of the function. The function $y = (x^2 + 9) \ln(x^2 + 9)$ is a product of two functions: - $u(x) = x^2 + 9$ - $v(x) = \ln(x^2 + 9)$

STEP 4

Apply the product rule for differentiation, which states that if $y = u(x) \cdot v(x)$ , then:
$y' = u'(x) \cdot v(x) + u(x) \cdot v'(x)$

STEP 5

Differentiate $u(x) = x^2 + 9$ :
$u'(x) = \frac{d}{dx}(x^2 + 9) = 2x$

STEP 6

Differentiate $v(x) = \ln(x^2 + 9)$ using the chain rule:
$v'(x) = \frac{d}{dx}[\ln(x^2 + 9)] = \frac{1}{x^2 + 9} \cdot \frac{d}{dx}(x^2 + 9) = \frac{1}{x^2 + 9} \cdot 2x = \frac{2x}{x^2 + 9}$

STEP 7

Substitute $u'(x)$ , $v(x)$ , $u(x)$ , and $v'(x)$ into the product rule formula:
$y' = (2x) \cdot \ln(x^2 + 9) + (x^2 + 9) \cdot \frac{2x}{x^2 + 9}$

STEP 8

Simplify the expression:
$y' = 2x \ln(x^2 + 9) + 2x$
The derivative of the function is:
$y' = 2x \ln(x^2 + 9) + 2x$

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QuestionDifferentiate the function y=(x2+9)ln⁡(x2+9)y=\left(x^{2}+9\right) \ln \left(x^{2}+9\right)y=(x2+9)ln(x2+9) y′=y^{\prime}=y′=

Studdy Solution

STEP 1

1. We are given the function y=(x2+9)ln⁡(x2+9) y = (x^2 + 9) \ln(x^2 + 9) y=(x2+9)ln(x2+9).2. We need to find the derivative y′ y' y′ with respect to x x x.

STEP 2

1. Identify the structure of the function and determine the differentiation method.2. Apply the product rule to differentiate the function.3. Simplify the derivative expression.

STEP 3

Identify the structure of the function. The function y=(x2+9)ln⁡(x2+9) y = (x^2 + 9) \ln(x^2 + 9) y=(x2+9)ln(x2+9) is a product of two functions: - u(x)=x2+9 u(x) = x^2 + 9 u(x)=x2+9 - v(x)=ln⁡(x2+9) v(x) = \ln(x^2 + 9) v(x)=ln(x2+9)

STEP 4

Apply the product rule for differentiation, which states that if y=u(x)⋅v(x) y = u(x) \cdot v(x) y=u(x)⋅v(x), then:y′=u′(x)⋅v(x)+u(x)⋅v′(x) y' = u'(x) \cdot v(x) + u(x) \cdot v'(x) y′=u′(x)⋅v(x)+u(x)⋅v′(x)

STEP 5

Differentiate u(x)=x2+9 u(x) = x^2 + 9 u(x)=x2+9:u′(x)=ddx(x2+9)=2x u'(x) = \frac{d}{dx}(x^2 + 9) = 2x u′(x)=dxd​(x2+9)=2x

STEP 6

STEP 7

Substitute u′(x) u'(x) u′(x), v(x) v(x) v(x), u(x) u(x) u(x), and v′(x) v'(x) v′(x) into the product rule formula:y′=(2x)⋅ln⁡(x2+9)+(x2+9)⋅2xx2+9 y' = (2x) \cdot \ln(x^2 + 9) + (x^2 + 9) \cdot \frac{2x}{x^2 + 9} y′=(2x)⋅ln(x2+9)+(x2+9)⋅x2+92x​

STEP 8

Simplify the expression:y′=2xln⁡(x2+9)+2x y' = 2x \ln(x^2 + 9) + 2x y′=2xln(x2+9)+2x The derivative of the function is:y′=2xln⁡(x2+9)+2x y' = 2x \ln(x^2 + 9) + 2x y′=2xln(x2+9)+2x

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QuestionDifferentiate the function $y=\left(x^{2}+9\right) \ln \left(x^{2}+9\right)$ $y^{\prime}=$

1. We are given the function $y = (x^2 + 9) \ln(x^2 + 9)$ .
2. We need to find the derivative $y'$ with respect to $x$ .

1. Identify the structure of the function and determine the differentiation method.
2. Apply the product rule to differentiate the function.
3. Simplify the derivative expression.

Identify the structure of the function. The function $y = (x^2 + 9) \ln(x^2 + 9)$ is a product of two functions: - $u(x) = x^2 + 9$ - $v(x) = \ln(x^2 + 9)$

Apply the product rule for differentiation, which states that if $y = u(x) \cdot v(x)$ , then:
$y' = u'(x) \cdot v(x) + u(x) \cdot v'(x)$

Differentiate $u(x) = x^2 + 9$ :
$u'(x) = \frac{d}{dx}(x^2 + 9) = 2x$

Substitute $u'(x)$ , $v(x)$ , $u(x)$ , and $v'(x)$ into the product rule formula:
$y' = (2x) \cdot \ln(x^2 + 9) + (x^2 + 9) \cdot \frac{2x}{x^2 + 9}$

Simplify the expression:
$y' = 2x \ln(x^2 + 9) + 2x$
The derivative of the function is:
$y' = 2x \ln(x^2 + 9) + 2x$