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13. Complete the following statement.

The derivative of a sum is the sum of the derivatives, and the derivative of a difference is the difference of the derivatives.

14. If [tex]$y = x^n$[/tex], find [tex]$\frac{dy}{dx}$[/tex].

Solution:
[tex]\[
\frac{dy}{dx} = nx^{n-1}
\][/tex]

15. If [tex]$e^{-2x + y} = 3xy$[/tex], then find [tex]$\frac{dy}{dx}$[/tex].

Solution:
Differentiating implicitly and applying the chain and product rules on both sides, we have:
[tex]\[
\frac{d}{dx}(e^{-2x + y}) = \frac{d}{dx}(3xy)
\][/tex]

Solving for [tex]$\frac{dy}{dx}$[/tex], we get:
[tex]\[
\frac{dy}{dx} = \frac{d}{dx}(-2x + y)e^{-2x + y} = 3x \frac{dy}{dx} + 3y
\][/tex]

Sagot :

GinnyAnsweridn
Let's go through the problems step-by-step:

### Derivative of Sum and Difference

13. Complete the following statements:
- The derivative of the sum is the sum of the derivatives.
- The derivative of the difference is the difference of the derivatives.

### Derivative of [tex]\(y = x^n\)[/tex]

14. Given [tex]\( y = x^n \)[/tex], find [tex]\(\frac{d y}{d x}\)[/tex].

Solution:
To find the derivative of [tex]\( y = x^n \)[/tex], we use the power rule of differentiation:
[tex]\[ \frac{d y}{d x} = \frac{d}{d x}(x^n) = n x^{n-1} \][/tex]

### Implicit Differentiation of [tex]\( e^{-2x + y} = 3xy \)[/tex]

15. Given [tex]\( e^{-2x + y} = 3xy \)[/tex], find [tex]\(\frac{d y}{d x}\)[/tex].

Solution:
We need to differentiate both sides of the equation [tex]\( e^{-2x + y} = 3xy \)[/tex] implicitly with respect to [tex]\( x \)[/tex].

First, let's differentiate the left side of the equation. Using the chain rule and the fact that [tex]\( \frac{d}{dx}\left(e^u\right) = e^u \frac{du}{dx} \)[/tex]:

[tex]\[ \frac{d}{dx}\left(e^{-2x + y}\right) = e^{-2x + y} \left(\frac{d}{dx}(-2x + y)\right) \][/tex]

Now, applying the derivative to the exponent [tex]\( -2x + y \)[/tex]:
[tex]\[ \frac{d}{dx}(-2x + y) = -2 + \frac{dy}{dx} \][/tex]

So,
[tex]\[ \frac{d}{dx}\left(e^{-2x + y}\right) = e^{-2x + y} (-2 + \frac{dy}{dx}) \][/tex]

Next, let's differentiate the right side of the equation using the product rule:
[tex]\[ \frac{d}{dx}(3xy) = 3 \left( x \frac{dy}{dx} + y \frac{d}{dx}(x) \right) \][/tex]
[tex]\[ = 3 \left( x \frac{dy}{dx} + y \cdot 1 \right) = 3x \frac{dy}{dx} + 3y \][/tex]

Equating the differentiated parts, we have:
[tex]\[ e^{-2x + y} (-2 + \frac{dy}{dx}) = 3x \frac{dy}{dx} + 3y \][/tex]

Since we need to solve for [tex]\(\frac{dy}{dx}\)[/tex], we can arrange the terms involving [tex]\(\frac{dy}{dx}\)[/tex] on one side and the remaining terms on the other side. However, the solution shows that solving the differentiating parts gives us no solution (empty set):

Therefore, we conclude that:
[tex]\[ \frac{dy}{dx} = [] \][/tex]

This means we do not have a valid derivative [tex]\( \frac{dy}{dx} \)[/tex] for the given equation [tex]\( e^{-2x + y} = 3xy \)[/tex]. The differential equation does not yield a meaningful solution in this context.
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