Derivatives: Learning Objective 4

Apply the Power Rule, $\frac{d}{dx}(x^n) = nx^{n-1}$, to each term of the polynomial. Remember that the derivative of a constant is zero.

$8x^7 + 60x^4 - 16x^3 + 30x^2 - 6$

We differentiate the polynomial term by term using the Power Rule: \[ \frac{d}{dx}(x^8 + 12x^5 - 4x^4 + 10x^3 - 6x + 5) \] \[ = \frac{d}{dx}(x^8) + 12\frac{d}{dx}(x^5) - 4\frac{d}{dx}(x^4) + 10\frac{d}{dx}(x^3) - 6\frac{d}{dx}(x) + \frac{d}{dx}(5) \] \[ = 8x^{8-1} + 12(5x^{5-1}) - 4(4x^{4-1}) + 10(3x^{3-1}) - 6(1) + 0 \] \[ = 8x^7 + 60x^4 - 16x^3 + 30x^2 - 6 \]

A horizontal tangent line occurs where the slope is zero. This means you need to find where the derivative of the function is equal to zero.

$(0, 4)$, $(\sqrt{3}, -5)$, and $(-\sqrt{3}, -5)$

1. Find the derivative of the function: \[ y' = \frac{d}{dx}(x^4 - 6x^2 + 4) = 4x^3 - 12x \] 2. Set the derivative to zero to find the x-coordinates: \[ 4x^3 - 12x = 0 \] \[ 4x(x^2 - 3) = 0 \] This gives us three possible values for $x$: $x=0$, $x=\sqrt{3}$, and $x=-\sqrt{3}$.

3. Find the corresponding y-coordinates:

• If $x=0$, then $y = (0)^4 - 6(0)^2 + 4 = 4$. Point: $(0,4)$.
• If $x=\sqrt{3}$, then $y = (\sqrt{3})^4 - 6(\sqrt{3})^2 + 4 = 9 - 6(3) + 4 = 9 - 18 + 4 = -5$. Point: $(\sqrt{3}, -5)$.
• If $x=-\sqrt{3}$, then $y = (-\sqrt{3})^4 - 6(-\sqrt{3})^2 + 4 = 9 - 6(3) + 4 = -5$. Point: $(-\sqrt{3}, -5)$.

You can solve this in two ways: either by using the Product Rule, or by first expanding the expression into a polynomial and then differentiating term-by-term.

$2x - 6x^2$

Method 1: Expand First
First, expand the function: $g(x) = x^2 - 2x^3$.
Now, differentiate using the Power Rule: \[ g'(x) = \frac{d}{dx}(x^2) - 2\frac{d}{dx}(x^3) = 2x - 2(3x^2) = 2x - 6x^2 \]

---

Method 2: Product Rule
Let $f(x)=x^2$ and $h(x)=1-2x$. Then $f'(x)=2x$ and $h'(x)=-2$. \[ g'(x) = f'(x)h(x) + f(x)h'(x) \] \[ = (2x)(1-2x) + (x^2)(-2) = 2x - 4x^2 - 2x^2 = 2x - 6x^2 \]

It's easiest to first simplify the expression. Divide each term in the numerator by the denominator, $\sqrt{x} = x^{1/2}$, and then use the Power Rule on the resulting terms.

$y' = \frac{3}{2}x^{1/2} + 2x^{-1/2} - \frac{3}{2}x^{-3/2}$

1. Simplify the function: \[ y = \frac{x^2}{x^{1/2}} + \frac{4x}{x^{1/2}} + \frac{3}{x^{1/2}} = x^{3/2} + 4x^{1/2} + 3x^{-1/2} \] 2. Differentiate using the Power Rule: \[ y' = \frac{d}{dx}(x^{3/2}) + 4\frac{d}{dx}(x^{1/2}) + 3\frac{d}{dx}(x^{-1/2}) \] \[ = \frac{3}{2}x^{(3/2)-1} + 4\left(\frac{1}{2}x^{(1/2)-1}\right) + 3\left(-\frac{1}{2}x^{(-1/2)-1}\right) \] \[ = \frac{3}{2}x^{1/2} + 2x^{-1/2} - \frac{3}{2}x^{-3/2} \]

You can use the Chain Rule, or more simply, expand the expression first and then differentiate.

$2a + 2$

Method 1: Expand First
Expand the expression: $R(a) = (a+1)^2 = a^2 + 2a + 1$.
Differentiate: $R'(a) = 2a + 2 + 0 = 2a + 2$.

Method 2: Chain Rule
Let $u = a+1$. Then $R(u) = u^2$. \[ R'(a) = \frac{dR}{du} \cdot \frac{du}{da} = (2u) \cdot (1) = 2(a+1) = 2a+2 \]

Rewrite the function using a negative exponent, $F(r) = 5r^{-3}$, and then apply the Power Rule.

$-\frac{15}{r^4}$

First, rewrite the function with a negative exponent: \[ F(r) = 5r^{-3} \] Now apply the Power Rule: \[ F'(r) = 5(-3)r^{-3-1} = -15r^{-4} = -\frac{15}{r^4} \]

Find the slope of the tangent line by evaluating the derivative at $x=1$. Then use the point-slope form, $y-y_1 = m(x-x_1)$, with the given point $(1,2)$.

$y=3x-1$

1. Find the derivative: \[ y' = \frac{d}{dx}(3x^2-x^3) = 6x - 3x^2 \] 2. Find the slope at the given point: Evaluate the derivative at $x=1$. \[ m = y'(1) = 6(1) - 3(1)^2 = 6 - 3 = 3 \] 3. Use the point-slope form: Using the point $(x_1, y_1) = (1, 2)$ and the slope $m=3$: \[ y - y_1 = m(x-x_1) \] \[ y - 2 = 3(x - 1) \] \[ y - 2 = 3x - 3 \] \[ y = 3x - 1 \]

Apply the Power Rule to find the first derivative. Then, differentiate the first derivative to find the second derivative.

$y' = 2 - \frac{15}{4}x^{-1/4}$
$y'' = \frac{15}{16}x^{-5/4}$

First Derivative: \[ y' = \frac{d}{dx}(2x - 5x^{3/4}) = 2 - 5\left(\frac{3}{4}x^{(3/4)-1}\right) \] \[ = 2 - \frac{15}{4}x^{-1/4} \]
Second Derivative: \[ y'' = \frac{d}{dx}\left(2 - \frac{15}{4}x^{-1/4}\right) = 0 - \frac{15}{4}\left(-\frac{1}{4}x^{(-1/4)-1}\right) \] \[ = \left(-\frac{15}{4}\right)\left(-\frac{1}{4}\right)x^{-5/4} = \frac{15}{16}x^{-5/4} \]