Practice Problem 1

Evaluate the following limit:

$\Huge \lim_{x \to 7} \frac{(x+1)^{1/3} - 2}{x-7}$

Answer: If we plug in $x = 7$, we see we get a divide by zero. So we can't just do direct substitution. Instead, we recognize this as a difference quotient that defines the derivative at some point

$\Huge \lim_{x \to a} \frac{f(x) - f(a)}{x-a}$

Reverse engineering the above formula, we see that $f(x) = (x+1)^{1/3}$ and $a = 7$, so this limit can be found as $f'(7)$. So now we need to compute $f'(x)$. This can be found by the chain rule to be

$\Large f'(x) = \frac{1}{3}(x+1)^{-2/3}$


$\Large f'(7) = \frac{1}{3}(8^{-2/3}) = \frac{1}{3 (\sqrt[3]{8})^2} = \frac{1}{12}$

Just to check that this seems right, let's compare it to a direct substitution with values very close to $7$:

In [8]:
f = ((x+1)^(1/3)-2)/(x-7)
In [7]:

Practice Problem 2

What is $\Large dy/dx$ at a point (x, y) on the curve $\Large e^{2y} = x^2 + y^2$?

Since we're taking the derivative of $y$ with respect to $x$, any time we see an expression with $x$, we can take the derivative directly, but when we see an expression with respect to $y$, we have to take the derivative with respect to $y$ and multiply by $dy/dx$. Applying that to the above equation when taking the derivative of both sides, we get

$\Large 2e^{2y} dy/dx = 2x + 2y dy/dx$

Re-arranging, we get

$\Large dy/dx (2e^{2y} - 2y) = 2x$

$\Large dy/dx = \frac{2x}{2e^{2y}-2y}$

Now let's try doing this the other way and computing $dx/dy$. In this case, it is the opposite; whenever we see a $y$, we take the derivative directly, and whenever we see an $x$, we take the derivative with respect to $x$, and mutiply by $dx/dy$. Doing this to both sides of the equation, we get

$\Large 2e^{2y} = 2x dx/dy + 2y$

$\Large dx/dy = \frac{2e^{2y} - 2y}{2x}$

Notice that this is just $1/(dy/dx)$! So if it's easier, you can just compute $dx/dy$ and take the reciprocal to get $dy/dx$.

Practice Problem 3

Evaluate the following limit (NOTE: the notation $\cos^2(x) = (\cos(x))^2$:

$\Large \lim_{x \to 0} \frac{\tan(x^2 + 2x)}{x}$

We notice that this leads to a 0/0 situation. But we should also recognize this as a difference quotient, where $f(x) = \tan(x^2 + 2x)$ and $a = 0$. Therefore, we should find the derivative $f'(x)$ and plug in 0 as $f'(0)$. In this case, we need the chain rule. The inner function is $x^2 + 2x$, and the outer function is $\tan(x)$. The derivative of the inner function is $2x + 2$, and the derivative of the outer function is $\sec^2(x)$. Therefore

$\Large f'(x) = \sec^2(x^2 + 2x)*(2x + 2) = \frac{2x+2}{\cos^2(x^2 + 2x)} = \frac{2}{1} = 2$

Let's check this below by plugging in numbers very close to zero

In [11]:
f = tan(x^2+2*x)/x
In [ ]: