Derivatives: Learning Objective 5

Use the Product Rule: $(uv)' = u'v + uv'$. Let $u=x$ and $v=e^x$.

$e^x(x+1)$

Using the Product Rule with $u=x$ and $v=e^x$, we have $u'=1$ and $v'=e^x$. \[ f'(x) = \frac{d}{dx}(xe^x) = (\frac{d}{dx}x)e^x + x(\frac{d}{dx}e^x) \] \[ = (1)e^x + x(e^x) = e^x + xe^x = e^x(1+x) \]

Use the Product Rule to find a general expression for $f'(x)$. Then, substitute $x=4$ and use the given values for $g(4)$ and $g'(4)$.

6.5

First, find the derivative using the Product Rule. Let $u=\sqrt{x}=x^{1/2}$ and $v=g(x)$. \[ f'(x) = (\frac{d}{dx}x^{1/2})g(x) + \sqrt{x}g'(x) \] \[ = \frac{1}{2}x^{-1/2}g(x) + \sqrt{x}g'(x) = \frac{g(x)}{2\sqrt{x}} + \sqrt{x}g'(x) \] Now, substitute $x=4$: \[ f'(4) = \frac{g(4)}{2\sqrt{4}} + \sqrt{4}g'(4) \] Use the given values $g(4)=2$ and $g'(4)=3$: \[ f'(4) = \frac{2}{2(2)} + 2(3) = \frac{2}{4} + 6 = 0.5 + 6 = 6.5 \]

Use the Quotient Rule: $(\frac{u}{v})' = \frac{u'v - uv'}{v^2}$.

$y' = \frac{-x^4-2x^3+6x^2+12x+6}{(x^3+6)^2}$

Let $u = x^2+x-2$ and $v = x^3+6$.
Then $u' = 2x+1$ and $v' = 3x^2$.

Applying the Quotient Rule: \[ y' = \frac{u'v - uv'}{v^2} = \frac{(2x+1)(x^3+6) - (x^2+x-2)(3x^2)}{(x^3+6)^2} \] Expand the numerator: \[ = \frac{(2x^4+12x+x^3+6) - (3x^4+3x^3-6x^2)}{(x^3+6)^2} \] \[ = \frac{2x^4+x^3+12x+6 - 3x^4-3x^3+6x^2}{(x^3+6)^2} \] Combine like terms: \[ = \frac{-x^4-2x^3+6x^2+12x+6}{(x^3+6)^2} \]

Find the slope by evaluating the derivative at $x=1$. Then use the point-slope form of a line.

$y = \frac{e}{2}$ (a horizontal line).

1. Find the derivative using the Quotient Rule: Let $u=e^x, v=1+x^2$. Then $u'=e^x, v'=2x$. \[ y' = \frac{e^x(1+x^2) - e^x(2x)}{(1+x^2)^2} = \frac{e^x(1+x^2-2x)}{(1+x^2)^2} = \frac{e^x(1-x)^2}{(1+x^2)^2} \] 2. Find the slope at $x=1$: \[ m = y'(1) = \frac{e^1(1-1)^2}{(1+1^2)^2} = \frac{e(0)^2}{2^2} = 0 \] 3. Use the point-slope form: A slope of 0 means the tangent line is horizontal. The equation of a horizontal line is $y=c$, where $c$ is the y-coordinate of the point. The equation is $y = \frac{e}{2}$.

Apply the Quotient Rule.

$\frac{e^{x}}{(1+e^x)^2}$

Let $u=e^x, v=1+e^x$. Then $u'=e^x, v'=e^x$. \[ f'(x) = \frac{u'v-uv'}{v^2} = \frac{e^x(1+e^x) - e^x(e^x)}{(1+e^x)^2} \] \[ = \frac{e^x+e^{2x}-e^{2x}}{(1+e^x)^2} = \frac{e^x}{(1+e^x)^2} \]

Simplify the complex fraction first by multiplying the numerator and denominator by $x$. Then apply the Quotient Rule.

$\frac{2cx}{(x^2+c)^2}$

1. Simplify the function: \[ f(x) = \frac{x}{x+\frac{c}{x}} \cdot \frac{x}{x} = \frac{x^2}{x^2+c} \] 2. Apply the Quotient Rule: Let $u=x^2, v=x^2+c$. Then $u'=2x, v'=2x$. \[ f'(x) = \frac{(2x)(x^2+c) - (x^2)(2x)}{(x^2+c)^2} \] \[ = \frac{2x^3+2cx - 2x^3}{(x^2+c)^2} = \frac{2cx}{(x^2+c)^2} \]

Use the Product Rule. Remember that $\sqrt{x} = x^{1/2}$.

$e^x(x + 2\sqrt{x} + 1 + x^{-1/2})$

Let $u=x+2x^{1/2}$ and $v=e^x$.
Then $u' = 1 + 2(\frac{1}{2}x^{-1/2}) = 1 + x^{-1/2}$. And $v'=e^x$.

Using the Product Rule: \[ g'(x) = u'v + uv' = (1+x^{-1/2})e^x + (x+2\sqrt{x})e^x \] \[ = e^x(1 + x^{-1/2} + x + 2\sqrt{x}) \]

Apply the Quotient Rule. You will need to use the Product Rule to find the derivative of the numerator, $x^2e^x$.

$y' = \frac{x^4e^x + 2xe^{2x}}{(x^2+e^x)^2}$

Let $u = x^2e^x$ and $v = x^2+e^x$.
$u' = (2x)e^x + x^2e^x = e^x(2x+x^2)$.
$v' = 2x+e^x$.

\[ y' = \frac{u'v - uv'}{v^2} = \frac{e^x(2x+x^2)(x^2+e^x) - (x^2e^x)(2x+e^x)}{(x^2+e^x)^2} \] Expand the numerator: \[ = \frac{e^x(2x^3+2xe^x+x^4+x^2e^x) - (2x^3e^x+x^2e^{2x})}{(x^2+e^x)^2} \] \[ = \frac{2x^3e^x+2xe^{2x}+x^4e^x+x^2e^{2x} - 2x^3e^x - x^2e^{2x}}{(x^2+e^x)^2} \] Combine like terms: \[ = \frac{x^4e^x + 2xe^{2x}}{(x^2+e^x)^2} \]

Use the Quotient Rule.

$h'(\theta) = \frac{-e^{-\theta}(2+\theta)}{(1+\theta)^2}$

Let $u=e^{-\theta}, v=1+\theta$. Then $u'=-e^{-\theta}, v'=1$. \[ h'(\theta) = \frac{u'v-uv'}{v^2} = \frac{(-e^{-\theta})(1+\theta) - (e^{-\theta})(1)}{(1+\theta)^2} \] \[ = \frac{-e^{-\theta}-\theta e^{-\theta}-e^{-\theta}}{(1+\theta)^2} = \frac{-2e^{-\theta}-\theta e^{-\theta}}{(1+\theta)^2} = \frac{-e^{-\theta}(2+\theta)}{(1+\theta)^2} \]

Use the Product Rule to differentiate $L(\lambda)$ with respect to $\lambda$. Treat $x_1$ and $x_2$ as constants. Let $u=\lambda^2$ and $v=e^{-\lambda(x_1+x_2)}$.

$\frac{d}{d\lambda}L(\lambda) = e^{-\lambda(x_1+x_2)}[2\lambda-\lambda^2(x_1+x_2)]$

Let $u=\lambda^2$ and $v=e^{-\lambda(x_1+x_2)}$. We are differentiating with respect to $\lambda$.
$u' = \frac{d}{d\lambda}(\lambda^2) = 2\lambda$.
$v' = \frac{d}{d\lambda}(e^{-\lambda(x_1+x_2)}) = e^{-\lambda(x_1+x_2)} \cdot (-(x_1+x_2))$ by the Chain Rule.

Using the Product Rule: \[ \frac{d}{d\lambda}L(\lambda) = u'v + uv' \] \[ = (2\lambda)(e^{-\lambda(x_1+x_2)}) + (\lambda^2)(-(x_1+x_2)e^{-\lambda(x_1+x_2)}) \] Factor out the common term $e^{-\lambda(x_1+x_2)}$: \[ = e^{-\lambda(x_1+x_2)}[2\lambda - \lambda^2(x_1+x_2)] \]