The derivative of a sum is the sum of the derivatives. Using the sum rule and the power rule we see. $$\begin{align*} \frac{d}{dx}\left(\sum_{k=1}^{21} kx^k\right) &=\sum_{k=1}^{21}\frac{d}{dx}\left(kx^k\right) \\ &=\sum_{k=1}^{21}k(kx^{k-1}) \\ &=\sum_{k=1}^{21}k^2x^{k-1}. \end{align*}$$

We use the chain rule.

We can write $f(x) = g(h(x))$ where $g(x) = \sqrt{x} = x^{1/2}$ and $h(x) = 4e^x+7x^2$. We then have $g'(x) = \frac{1}{2}x^{-1/2}$ and $h'(x) = 4e^x+14x$. The chain rule then gives $$f'(x) = g'(h(x))h'(x) = \frac{1}{2}(h(x))^{-1/2}(4e^x+14x) = \frac{1}{2}(4e^x+7x^2)^{-1/2}(4e^x+14x)$$.

• We have $\frac{df}{dx} = m$ given $b$ and $m$ are constant. Soon we will write this using partial derivative notation as $\frac{\partial f}{\partial x}= m$. The stylied $d$ written as $\partial$ means that you assume all other variables are constant.
• We have $\frac{df}{dm} = x$ given $x$ and $b$ are constant. Partial derivative notation gives $\frac{\partial f}{\partial m}= x$.
• We have $\frac{df}{db} = 1$ given $x$ and $m$ are constant. Partial derivative notation gives $\frac{\partial f}{\partial b} = 1$.

• We have $\frac{dg}{dx} = 2x+5(1)y-0$ given $y$ is constant. We can write this as $\frac{\partial g}{\partial x}=2x+5y$.
• We have $\frac{dg}{dy} = 0+5x(1)-3y^2$ given $x$ is constant. We can write this as $\frac{\partial g}{\partial y}=5x-3y^2$.

• We have $\frac{dg}{dx} = 3x^2y^3$ given $y$ is constant. We can write this as $\frac{\partial g}{\partial x}=3x^2y^3$.
• We have $\frac{dg}{dy} = 3x^3y^2 - 5$ given $x$ is constant. We can write this as $\frac{\partial g}{\partial y}=3x^3y^2 - 5$.

• We have $\frac{dh}{dx} = 5(x^2e^y - a\ln(x) - 10)^4(2x e^y - \frac{a}{x})$ given $y$ and $a$ are constant, or simply $\frac{\partial h}{\partial x} = 5(x^2e^y - a\ln(x) - 10)^4(2x e^y - \frac{a}{x})$.
• We have $\frac{dh}{dy} = 5(x^2e^y - a\ln(x) - 10)^4(x^2 e^y)$ given $x$ and $a$ are constant, or simply $\frac{\partial h}{\partial y} = 5(x^2e^y - a\ln(x) - 10)^4(x^2 e^y)$.
• We have $\frac{dh}{da} = 5(x^2e^y - a\ln(x) - 10)^4(-\ln(x))$ given $x$ and $y$ are constant, or simply $\frac{\partial h}{\partial a} = 5(x^2e^y - a\ln(x) - 10)^4(-\ln(x))$.

• $\frac{df}{dx}(x) = (\frac{d}{dx}(x^2 - 1))(-x^3 - 5x +2) + (x^2 - 1)\frac{d}{dx}(-x^3 - 5x +2) = (2x)(-x^3 - 5x +2) + (x^2 - 1)(-3x^2-5) = -5x^4 - 12x^2 + 4x + 5$
• $\frac{dh}{dx}(x) = \frac{ (\frac{d}{dx}(-2x^2 + x + 4))(x^2 + 4x -1) - (-2x^2 + x + 4)\frac{d}{dx}(x^2 + 4x -1)}{ (x^2 + 4x -1)^2} = \frac{ (-4x + 1)(x^2 + 4x -1) - (-2x^2 + x + 4)(2x + 4) }{ (x^2 + 4x -1)^2} = \frac{-9x^2 -4x-17}{ (x^2 + 4x -1)^2}$
• $\frac{dg}{dx}(x) = \frac{ (\frac{d}{dx}(1-2x)(x+4))(x^2 + 3x - 1) - (1-2x)(x+4)\frac{d}{dx}(x^2 + 3x - 1) }{ (x^2 + 3x - 1)^2 } = \frac{ ((\diff{}{x}(1-2x))(x+4) + (1-2x)\diff{}{x}(x+4))(x^2 + 3x - 1) - (1-2x)(x+4)(2x + 3) }{ (x^2 + 3x - 1)^2 } = \frac{ x^2 -4x -5 }{ (x^2 + 3x - 1)^2 }$
• $\frac{dy}{dx}(x) = 0$
• $\frac{ds}{dx}(x) = 12a^2x^{3a-1} - 2bcx^{2c-1}$
• $\frac{dr}{dx}(x) = -\frac{3}{2}(1-3x)^{-\frac{1}{2}} = -\frac{3}{2\sqrt{1-3x}}$
• $\frac{dl}{dx} = -3 + 14x$
• $\frac{dq}{dt} = -3e^t$
• $\frac{dz}{dx} = -\frac{2}{x}$
• $\frac{dw}{dx} = \frac{ 3x^2 + 22x - 25 }{ (3-x)^2(5x+2)^2 }$

Recall that when we find the derivative of a function $f(x)$ at a point $x=a$, we are find the slope of the linear approximation to the function at $x=a$. The graph of the linear approximation is a line that is tangent to the graph of the function. This activity focuses on finding the linear approximation to a function at a point, and writing an equation for it in various ways.

As an example, let's find the linear approximation to $f(x) = x^2$ at $x = 3$. The linear approximation must pass through the point $x=3$ and $y = f(3) = 3^2 = 9$. The derivative of $f$ is $f'(x) = 2x$. The slope of the linear approximation is $m=f'(3) = 2(3) = 6$. So the graph of the linear approximation is a line through the point $(x,y) =(3,9)$ with slope $f'(3)=6$.

• Using point-slope form, an equation for the tangent line is $y - 9 = 6(x-3)$.
• Solving for $y$ above gives $y = L(x) = 9+6(x-3)$. It's common to use the form $L(x) = f(a) + f'(a)(x-a)$ when stating a linear approximation.
• We can write the above in slope-intercept form to obtain $y = 6x-9$.

The point-slope form is useful for quickly writing an equation of a line when we know the point and slope. The form $L(x) = f(a) + f'(a)(x-a)$ preserves the fact that we wanted the linear approximation at $x=3$. The slope-intercept form is unique, and allows quick comparison. All three forms are useful.

1. 1. $y + 3 = -3x$
   2. $y = -3 - 3(x-0)$
   3. $y = -3 - 3x$
2. 1. $y - 172 = -45(x+3)$
   2. $y = L(x) = 172 -45(x+3)$
   3. $y = L(x) = 37 - 45x$
3. 1. $y - 1 = -2(x-1)$
   2. $y = L(x) = 1 -2(x-1)$
   3. $y = L(x) = 3 -2x$

First we write a formula for the residuals (or errors) to obtain $$r_i(A, b, k; t_i, c_i) = c_i - f(t_i; A, b, k) = c_i - (Ae^{bt_i} + k) = c_i - Ae^{bt_i} - k.$$

Because we are told the errors (or residuals) are normally distributed (with mean of 0 and standard deviation of 1), the probability model we will use is $$p(r) = \frac{1}{\sqrt{2\pi}}e^{-\frac{r^2}{2}}.$$ Or we could write this $p(A, b, k; t_i, c_i) = \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(c_i - Ae^{bt_i} - k)^2}.$

Because the errors are independent we can multiply the probability models to find the joint probability model as $$J(t_i, c_i;A, b, k) = \prod_{i=1}^{50} \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(c_i - Ae^{bt_i} - k)^2}.$$

The likelihood function (interchange the role of variables and parameters) is then $$L(A, b, k; t_i, c_i) = \prod_{i=1}^{50} \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(c_i - Ae^{bt_i} - k)^2}.$$

We take the natural log of the likelihood function to find the loglikelihood function as $$\ell(A, b, k; t_i, c_i) = \ln\left(\prod_{i=1}^{50} \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(c_i - Ae^{bt_i} - k)^2}\right).$$

We rewrite $\ell$, using the properties of logarithms and sums, to obtain $$\ell(A, b, k; t_i, c_i) = 50\ln\left(\frac{1}{\sqrt{2\pi}}\right) -\frac{1}{2}\sum_{i=1}^{50} (c_i - Ae^{bt_i} - k)^2.$$