Question

Math

Posted 5 months ago

```
Let $y=\frac{1-2 x}{3 x^{2}}$.
What is the value of $\frac{d y}{d x}$ at $x=1$ ?
```

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Answer from Sia

Posted 5 months ago

Solution by Steps

step 1

Given the function $y=\frac{1-2x}{3x^2}$, we need to find the derivative $\frac{dy}{dx}$

step 2

From the asksia-ll calculation list, the derivative of the function with respect to $x$ is $\frac{dy}{dx} = \frac{2(x-1)}{3x^3}$

step 3

To find the value of $\frac{dy}{dx}$ at $x=1$, we substitute $x=1$ into the derivative

step 4

Substituting $x=1$ gives $\frac{dy}{dx} = \frac{2(1-1)}{3(1)^3} = \frac{2(0)}{3} = 0$

Answer

$\frac{dy}{dx} = 0$ at $x=1$

Key Concept

Differentiation and evaluation at a point

Explanation

The value of the derivative of a function at a specific point is found by substituting the point into the derivative expression.

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