Here are the differentiations for each expression. 1. Differentiate $2x^2 + 3x + 4$. Step 1: Apply the sum rule of differentiation, which states that the derivative of a sum is the sum of the derivatives. $$ \frac{d}{dx}(2x^2 + 3x + 4) = \frac{d}{dx}(2x^2) + \frac{d}{dx}(3x) + \frac{d}{dx}(4) $$ Step 2: Apply the power rule $\frac{d}{dx}(ax^n) = anx^{n-1}$ and the constant rule $\frac{d}{dx}(c) = 0$. $$ \frac{d}{dx}(2x^2) = 2 \cdot 2x^{2-1} = 4x $$ $$ \frac{d}{dx}(3x) = 3 \cdot 1x^{1-1} = 3x^0 = 3 $$ $$ \frac{d}{dx}(4) = 0 $$ Step 3: Combine the results. $$ \frac{d}{dx}(2x^2 + 3x + 4) = 4x + 3 + 0 = 4x + 3 $$ The derivative is $\boxed{4x + 3}$. 2. Differentiate $\frac{7}{8}x^2 + 4x - 3$. Step 1: Apply the sum/difference rule of differentiation. $$ \frac{d}{dx}\left(\frac{7}{8}x^2 + 4x - 3\right) = \frac{d}{dx}\left(\frac{7}{8}x^2\right) + \frac{d}{dx}(4x) - \frac{d}{dx}(3) $$ Step 2: Apply the power rule and the constant rule. $$ \frac{d}{dx}\left(\frac{7}{8}x^2\right) = \frac{7}{8} \cdot 2x^{2-1} = \frac{14}{8}x = \frac{7}{4}x $$ $$ \frac{d}{dx}(4x) = 4 \cdot 1x^{1-1} = 4 $$ $$ \frac{d}{dx}(3) = 0 $$ Step 3: Combine the results. $$ \frac{d}{dx}\left(\frac{7}{8}x^2 + 4x - 3\right) = \frac{7}{4}x + 4 - 0 = \frac{7}{4}x + 4 $$ The derivative is $\boxed{\frac{7}{4}x + 4}$. 3. Differentiate $\frac{1}{(x^2 - 2)^3}$. Step 1: Rewrite the expression using a negative exponent. $$ y = \frac{1}{(x^2 - 2)^3} = (x^2 - 2)^{-3} $$ Step 2: Apply the chain rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$. Let $u = x^2 - 2$. Then $y = u^{-3}$. Step 3: Differentiate $y$ with respect to $u$. $$ \frac{dy}{du} = -3u^{-3-1} = -3u^{-4} $$ Step 4: Differentiate $u$ with respect to $x$. $$ \frac{du}{dx} = \frac{d}{dx}(x^2 - 2) = 2x - 0 = 2x $$ Step 5: Multiply the derivatives and substitute $u$ back. $$ \frac{dy}{dx} = (-3u^{-4})(2x) = -6x u^{-4} $$ Substitute $u = x^2 - 2$: $$ \frac{dy}{dx} = -6x(x^2 - 2)^{-4} $$ Step 6: Rewrite the expression with a positive exponent. $$ \frac{dy}{dx} = -\frac{6x}{(x^2 - 2)^4} $$ The derivative is $\boxed{-\frac{6x}{(x^2 - 2)^4}}$. 4. Differentiate $x^2 - 3x + 4$. The notation $\frac{d}{dx}(x^2 - 3x + 4)$ means to find the derivative of the expression with respect to $x$. Step 1: Apply the sum/difference rule of differentiation. $$ \frac{d}{dx}(x^2 - 3x + 4) = \frac{d}{dx}(x^2) - \frac{d}{dx}(3x) + \frac{d}{dx}(4) $$ Step 2: Apply the power rule and the constant rule. $$ \frac{d}{dx}(x^2) = 2x^{2-1} = 2x $$ $$ \frac{d}{dx}(3x) = 3 \cdot 1x^{1-1} = 3 $$ $$ \frac{d}{dx}(4) = 0 $$ Step 3: Combine the results. $$ \frac{d}{dx}(x^2 - 3x + 4) = 2x - 3 + 0 = 2x - 3 $$ The derivative is $\boxed{2x - 3}$. Send me the next one 📸