Let’s differentiate each function with respect to \( x \) step by step.
1. Differentiate \( 6x^5 + 4x^3 - 7x^2 + 3x - 8 \)
Using the power rule \( \frac{d}{dx}[x^n] = nx^{n-1} \):
\[ \frac{d}{dx}(6x^5) = 30x^4 \] \[ \frac{d}{dx}(4x^3) = 12x^2 \] \[ \frac{d}{dx}(-7x^2) = -14x \] \[ \frac{d}{dx}(3x) = 3 \] \[ \frac{d}{dx}(-8) = 0 \]
Combining all these results:
\[ \frac{d}{dx}(6x^5 + 4x^3 - 7x^2 + 3x - 8) = 30x^4 + 12x^2 - 14x + 3 \]
2. Differentiate \( (3x + 2)(6x + 5) \)
We will use the product rule: \( \frac{d}{dx}[u v] = u'v + uv' \), where \( u = 3x + 2 \) and \( v = 6x + 5 \).
First, find \( u' \) and \( v' \): \[ u' = \frac{d}{dx}(3x + 2) = 3 \] \[ v' = \frac{d}{dx}(6x + 5) = 6 \]
Now, apply the product rule: \[ \frac{d}{dx}[(3x + 2)(6x + 5)] = u'v + uv' = 3(6x + 5) + (3x + 2)(6) \]
Calculating the two terms: \[ 3(6x + 5) = 18x + 15 \] \[ (3x + 2)(6) = 18x + 12 \]
Combining these: \[ \frac{d}{dx}[(3x + 2)(6x + 5)] = (18x + 15) + (18x + 12) = 36x + 27 \]
3. Differentiate \( \frac{2x + 3}{3x + 3} \)
We will use the quotient rule: \( \frac{d}{dx}\left[\frac{u}{v}\right] = \frac{u'v - uv'}{v^2} \), where \( u = 2x + 3 \) and \( v = 3x + 3 \).
Finding \( u' \) and \( v' \): \[ u' = \frac{d}{dx}(2x + 3) = 2 \] \[ v' = \frac{d}{dx}(3x + 3) = 3 \]
Applying the quotient rule: \[ \frac{d}{dx}\left[\frac{2x + 3}{3x + 3}\right] = \frac{(2)(3x + 3) - (2x + 3)(3)}{(3x + 3)^2} \]
Simplifying the numerator: \[ 2(3x + 3) = 6x + 6 \] \[ (2x + 3)(3) = 6x + 9 \]
Putting it together: \[ \frac{d}{dx}\left[\frac{2x + 3}{3x + 3}\right] = \frac{(6x + 6) - (6x + 9)}{(3x + 3)^2} = \frac{6 - 9}{(3x + 3)^2} = \frac{-3}{(3x + 3)^2} \]
Final Results:
- \( 30x^4 + 12x^2 - 14x + 3 \)
- \( 36x + 27 \)
- \( \frac{-3}{(3x + 3)^2} \)