If `y=6 x^(5)-4x^(4)+2x^(2)+3x+2`, then find `(dy)/(dx)` at x=-1.

Answer

To find \(\frac{dy}{dx}\) for the function \(y = 6x^5 - 4x^4 + 2x^2 + 3x + 2\) at \(x = -1\), we will follow these steps:

### Step 1: Differentiate the function
We need to differentiate \(y\) with respect to \(x\). We will apply the power rule of differentiation, which states that \(\frac{d}{dx}(x^n) = n \cdot x^{n-1}\).

1. Differentiate \(6x^5\):
\[ \frac{d}{dx}(6x^5) = 6 \cdot 5 \cdot x^{5-1} = 30x^4 \]

2. Differentiate \(-4x^4\):
\[ \frac{d}{dx}(-4x^4) = -4 \cdot 4 \cdot x^{4-1} = -16x^3 \]

3. Differentiate \(2x^2\):
\[ \frac{d}{dx}(2x^2) = 2 \cdot 2 \cdot x^{2-1} = 4x \]

4. Differentiate \(3x\):
\[ \frac{d}{dx}(3x) = 3 \]

5. The derivative of the constant \(2\) is \(0\).

Combining all these, we get:
\[ \frac{dy}{dx} = 30x^4 - 16x^3 + 4x + 3 \]

### Step 2: Evaluate the derivative at \(x = -1\)
Now we will substitute \(x = -1\) into the derivative we found.

\[ \frac{dy}{dx} \bigg|_{x=-1} = 30(-1)^4 - 16(-1)^3 + 4(-1) + 3 \]

Calculating each term:
- \(30(-1)^4 = 30 \cdot 1 = 30\)
- \(-16(-1)^3 = -16 \cdot (-1) = 16\)
- \(4(-1) = -4\)
- The constant term is \(3\).

Putting it all together:
\[ \frac{dy}{dx} \bigg|_{x=-1} = 30 + 16 - 4 + 3 \]

Calculating the sum:
\[ 30 + 16 = 46 \]
\[ 46 - 4 = 42 \]
\[ 42 + 3 = 45 \]

Thus, the value of \(\frac{dy}{dx}\) at \(x = -1\) is:
\[ \frac{dy}{dx} \bigg|_{x=-1} = 45 \]

### Final Answer:
\[ \frac{dy}{dx} \text{ at } x = -1 \text{ is } 45. \]