(10 pts.) Find the tangent line to $y = \frac{1}{3}x^2$ at $x = 1$

Since $y' = \frac{2}{3}x$.

We have $\sout{y(0) = \frac{1}{3} \cdot 0^2 = 0}$ and $\sout{y'(0) = \frac{2}{3} \cdot 0 = 0}$.

Then $\sout{y - y(0) = y'(0)(x - 0) \Rightarrow y - 0 = 0 \cdot (x - 0) \Rightarrow y = 0}$.

So the tangent line is $\sout{y = 0}$.

We have $y(1) = \frac{1}{3} \cdot 1^2 = \frac{1}{3}$ and $y'(1) = \frac{2}{3} \cdot 1 = \frac{2}{3}$.

Then $y - y(1) = y'(1)(x - 1) \Rightarrow y - \frac{1}{3} = \frac{2}{3} \cdot (x - 1) \Rightarrow y = \frac{2}{3}x - \frac{1}{3}$.

Find the derivative of the following functions:

1. (7 pts.) $\frac{x}{\sqrt{1 - x}}$

   $(\frac{x}{\sqrt{1 - x}})' = (x(1-x)^{-\frac{1}{2}})' = (1-x)^{-\frac{1}{2}} + (x \cdot (-\frac{1}{2})(1-x)^{-\frac{3}{2}}) = \frac{1-\frac{\cancel{3} 1}{2}x}{(1-x)\sqrt{1-x}}$

2. (8 pts.) $\frac{\cos(2x)}{x}$

   $(\frac{\cos(2x)}{x})' = \frac{-2x \sin(2x) - \cos(2x)}{x^2}$

3. (5 pts.) $e^{2f(x)} = g(x)$

   $g'(x) = (e^{2f(x)})' = 2f'(x)e^{2f(x)}$

4. (5 pts.) $\ln(\sin x)$

   $(\ln(\sin x)) = \frac{1}{\sin x} \cdot \cos x = \cot x$

(15 pts.) Find $\frac{\mathrm{d}y}{\mathrm{d}x}$ for the function for the function $y$ defined implicitly by $y^4 + xy = 4$ at $x = 3, y = 1$.

$\frac{\mathrm{d}}{\mathrm{d}x}(y^4 + xy) = (4y^3)y' + y + xy' = y'(4y^3 + x) + y$

Since $y^4 + xy = 4$ and $\frac{\mathrm{d}}{\mathrm{d}x} 4 = 0$.

We have $y'(4y^3 + x) + y = 0$

Then $\frac{\mathrm{d}y}{\mathrm{d}x} = y' = -\frac{y}{4y^3 + x} = -\frac{1}{4 \cdot 1^3 + 3} = -\frac{1}{7}$ at $x = 3, y = 1$.

(15 pts.) Skipped.

(15 pts.) Let

Find all $a$ and $b$ such that the function $f(x)$ is differentiable.

So we have

at $x = 1$.

Then

And we can get $a = 5, b = -2$.

Evaluate these limits by relating them to a derivative.

1. (5 pts.) Evaluate $\lim_{x \rightarrow 0} \frac{(1 + 2x)^{10}-1}{x}$

   $\lim_{x \rightarrow 0} \frac{(1 + 2x)^{10}-1}{x} = 2 \lim_{x \rightarrow 0} \frac{(1 + 2x)^{10}-1^{10}}{2x} = 2 (t^{10})'|_{t = 1} = 20 t^9|_{t = 1} = 20$

2. (5 pts.) Evaluate $\lim_{x \rightarrow 0} \frac{\sqrt{\cos x}-1}{x}$

   $\lim_{x \rightarrow 0} \frac{\sqrt{\cos x}-1}{x} =\lim_{x \rightarrow 0} \frac{\sqrt{\cos (0 + x)}-\sqrt{\cos 0}}{x} =(\sqrt{\cos t})'|_{t = 0} =(-\frac{\sin t}{2\sqrt{\cos t}})|_{t = 0} =0$

(10 pts.) Derive the formula $\frac{\mathrm{d}}{\mathrm{d}x} a^x = M(a)a^x$ directly from the definition of the derivative, and identity $M(a)$ as a limit.

where $M(a) = \lim_{\Delta x \rightarrow 0}\frac{a^{\Delta x} - 1}{\Delta x}$