Let `H(y)=2y^(4)-6y^(3)+2y-4`. Find `H^(')(2)`.

Answer

To find \( H'(2) \) for the function \( H(y) = 2y^4 - 6y^3 + 2y - 4 \), we will first need to differentiate the function \( H(y) \) with respect to \( y \) and then evaluate the derivative at \( y = 2 \).

### Step 1: Differentiate \( H(y) \)

Using the power rule, which states that if \( h(y) = y^n \), then \( h'(y) = n \cdot y^{n-1} \), we can differentiate each term of \( H(y) \):

1. Differentiate \( 2y^4 \):
\[ \frac{d}{dy}(2y^4) = 2 \cdot 4y^{4-1} = 8y^3 \]

2. Differentiate \( -6y^3 \):
\[ \frac{d}{dy}(-6y^3) = -6 \cdot 3y^{3-1} = -18y^2 \]

3. Differentiate \( 2y \):
\[ \frac{d}{dy}(2y) = 2 \]

4. Differentiate \( -4 \):
\[ \frac{d}{dy}(-4) = 0 \]

Now, combining these results, we have:
\[ H'(y) = 8y^3 - 18y^2 + 2 \]

### Step 2: Evaluate \( H'(2) \)

Now we will substitute \( y = 2 \) into \( H'(y) \):
\[ H'(2) = 8(2^3) - 18(2^2) + 2 \]

Calculating each term:
1. \( 2^3 = 8 \) so \( 8(2^3) = 8 \times 8 = 64 \)
2. \( 2^2 = 4 \) so \( -18(2^2) = -18 \times 4 = -72 \)
3. The constant term is \( +2 \)

Now, substituting these values into the equation:
\[ H'(2) = 64 - 72 + 2 \]
\[ H'(2) = 64 - 72 = -8 \]
\[ H'(2) = -8 + 2 = -6 \]

### Final Answer
Thus, \( H'(2) = -6 \).

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