Math / Calculus

Question $\sqrt{y} - \ln y = x + 8$
Find $\frac{dy}{du}$ .

Studdy Solution

STEP 1

1. The equation $\sqrt{y} - \ln y = x + 8$ involves both algebraic and logarithmic functions.
2. We need to find the derivative $\frac{dy}{du}$ , which implies that $y$ is a function of $u$ .
3. We assume $x$ is a function of $u$ , i.e., $x = x(u)$ .

STEP 2

1. Differentiate the equation with respect to $u$ .
2. Solve for $\frac{dy}{du}$ .

STEP 3

Differentiate both sides of the equation $\sqrt{y} - \ln y = x + 8$ with respect to $u$ .
For the left side: - The derivative of $\sqrt{y}$ with respect to $u$ is $\frac{1}{2\sqrt{y}} \cdot \frac{dy}{du}$ . - The derivative of $-\ln y$ with respect to $u$ is $-\frac{1}{y} \cdot \frac{dy}{du}$ .
For the right side: - The derivative of $x$ with respect to $u$ is $\frac{dx}{du}$ . - The derivative of the constant $8$ is $0$ .
Putting it all together, we have:
$\frac{1}{2\sqrt{y}} \cdot \frac{dy}{du} - \frac{1}{y} \cdot \frac{dy}{du} = \frac{dx}{du}$

STEP 4

Factor out $\frac{dy}{du}$ from the left side of the equation:
$\left(\frac{1}{2\sqrt{y}} - \frac{1}{y}\right) \cdot \frac{dy}{du} = \frac{dx}{du}$

STEP 5

Solve for $\frac{dy}{du}$ by dividing both sides by $\left(\frac{1}{2\sqrt{y}} - \frac{1}{y}\right)$ :
$\frac{dy}{du} = \frac{\frac{dx}{du}}{\frac{1}{2\sqrt{y}} - \frac{1}{y}}$
The derivative $\frac{dy}{du}$ is:
$\frac{dy}{du} = \frac{\frac{dx}{du}}{\frac{1}{2\sqrt{y}} - \frac{1}{y}}$

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