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Find the derivative of $f(x) = x^n$ where $n$ is a positive integer.
$\displaystyle{\begin{array}{l} \displaystyle{\lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}} \\\\ \quad = \displaystyle{\lim_{h \rightarrow 0} \frac{(x+h)^n - x^n}{h}} \\\\ \quad = \displaystyle{\lim_{h \rightarrow 0} \frac{(x^n + nx^{n-1}h + ({}_n C_2)x^{n-2}h^2 + \cdots + ({}_n C_k)x^{n-k}h^k + \cdots + h^n) - x^n}{h}} \\ \quad \quad \quad \quad \textrm{(by the binomial theorem)}\\\\ \quad = \displaystyle{\lim_{h \rightarrow 0} \frac{nx^{n-1}h + ({}_n C_2)x^{n-2}h^2 + \cdots + ({}_n C_k)x^{n-k}h^k + \cdots + h^n}{h}}\\\\ \quad = \displaystyle{\lim_{h \rightarrow 0} \,\, [nx^{n-1} + ({}_n C_2)x^{n-2}h + \cdots + ({}_n C_k)x^{n-k}h^{k-1} + \cdots + h^{n-1}]}\\\\ \quad = \displaystyle{nx^{n-1} + 0 + \cdots + 0}\\\\ \quad = nx^{n-1} \end{array}}$
If $f'(x)$ is defined, find the derivative of $\displaystyle{\frac{1}{f(x)}}$.
This is an interesting question in that it seeks a formula, presumably in terms of $f'(x)$, for the derivative of $\displaystyle{\frac{1}{f(x)}}$
Towards this end, we will want to find some starting expression for $\displaystyle{\frac{1}{f(x)}}$ and try to "massage it" into revealing the limit in the definition of $f'(x)$ inside it somewhere.
As such, we first find the derivative of $\displaystyle{\frac{1}{f(x)}}$. We'll use Leibniz notation, as this function was not given an explicit name.
$$\frac{d}{dx} \left( \frac{1}{f(x)} \right) = \lim_{h \rightarrow 0} \frac{\displaystyle{\frac{1}{f(x+h)} - \frac{1}{f(x)}}}{h}$$Now, recall that
$$f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$We need to manipulate the first limit so that we can see the second limit inside of it. Notice the pieces are all there -- we have a $f(x+h)$, $f(x)$, and $h$ all present. Let's follow our natural inclination to simplify the complex fraction (by multiplying top and bottom by $f(x+h)f(x)$) and see if that gets us anywhere...
$$\begin{array}{rcl} \displaystyle{\frac{d}{dx} \left( \frac{1}{f(x)} \right)} &=& \displaystyle{\lim_{h \rightarrow 0} \frac{\displaystyle{\frac{1}{f(x+h)} - \frac{1}{f(x)}}}{h}}\\\\ &=& \displaystyle{\lim_{h \rightarrow 0} \frac{f(x) - f(x+h)}{h f(x+h)f(x)}}\\\\ &=& \displaystyle{\lim_{h \rightarrow 0} \left[\frac{f(x+h)-f(x)}{h} \cdot \frac{-1}{f(x+h)f(x)}\right]}\\\\ \end{array}$$Upon seing the desired fraction, we now use the limit law for the limit of a product to isolate the $f'(x)$ part, and then simplify the rest of the expression.
$$\begin{array}{rcl} &=& \displaystyle{\left[ \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \right] \cdot \left[ \lim_{h \rightarrow 0} \frac{-1}{f(x+h)f(x)} \right]}\\\\ &=& \displaystyle{f'(x) \cdot \left[ \lim_{h \rightarrow 0} \frac{-1}{f(x+h)f(x)} \right]}\\\\ &=& \displaystyle{f'(x) \cdot \frac{-1}{(f(x))^2}}\\\\ &=& \displaystyle{-\frac{f'(x)}{(f(x))^2}} \end{array}$$So the derivative of $\displaystyle{\frac{1}{f(x)}}$ in terms of $f'(x)$ is given by $\displaystyle{-\frac{f'(x)}{(f(x))^2}}$.
Ta-Da! 😀
Find the equation of the tangent line to $y=1-2\sqrt{4-x}$ at the point where $x=2$.
We could tackle this problem in a couple of ways. One way to proceed is to find the derivative of this function, which can then be used to tell us the slope of the tangent line at $x=2$.
This, in combination with the coordinates of the point on the graph at $x=2$, provides us enough information to find the equation of the tangent line.
$$\begin{array}{rcl} \displaystyle{\frac{dy}{dx}} &=& \displaystyle{\lim_{h \rightarrow 0} \frac{(1-2\sqrt{4-(x+h)}) - (1-2\sqrt{4-x})}{h}}\\\\ &=& \displaystyle{\lim_{h \rightarrow 0} \frac{-2\sqrt{4-x-h} + 2\sqrt{4-x}}{h}}\\\\ &=& \displaystyle{-2 \lim_{h \rightarrow 0} \frac{\sqrt{4-x-h} - \sqrt{4-x}}{h} \quad \textrm{ ...now use the conjugate trick}}\\\\ &=& \displaystyle{-2 \lim_{h \rightarrow 0} \left[ \frac{(\sqrt{4-x-h} - \sqrt{4-x})}{h} \cdot \frac{(\sqrt{4-x-h} + \sqrt{4-x})}{(\sqrt{4-x-h} + \sqrt{4-x})} \right]}\\\\ &=& \displaystyle{-2 \lim_{h \rightarrow 0} \frac{(4-x-h) - (4-x)}{h(\sqrt{4-x-h} + \sqrt{4-x})}}\\\\ &=& \displaystyle{-2 \lim_{h \rightarrow 0} \frac{-h}{h(\sqrt{4-x-h} + \sqrt{4-x})}}\\\\ &=& \displaystyle{-2 \lim_{h \rightarrow 0} \frac{-1}{(\sqrt{4-x-h} + \sqrt{4-x})}}\\\\ &=& \displaystyle{-2 \cdot \frac{-1}{2\sqrt{4-x}}}\\\\ &=& \displaystyle{\frac{1}{\sqrt{4-x}}} \end{array}$$Then note that at $x=2$, the slope f the tangent line is given by $\displaystyle{f'(2) = \frac{1}{\sqrt{2}}}$.
Note also, for $x=2$, $y=1-2\sqrt{2}$, giving us a point on the tangent line of $(2,1-2\sqrt{2})$.
From there, we can simply use the point-slope formula for a line to get the equation of the tangent line:
$$y - (1-2\sqrt{2}) = \frac{1}{\sqrt{2}} (x - 2)$$Another, slightly quicker way to work this problem -- since we only care about the tangent line at $x=2$ -- is to set up the limit for the derivative just for this value.
In other words, we find the tangent slope given by the following limit
$$\begin{array}{l} \displaystyle{\lim_{x \rightarrow 2} \frac{(1-2\sqrt{4-x}) - (1-2\sqrt{2})}{x-2}}\\\\ \quad = \displaystyle{\lim_{x \rightarrow 2} \frac{2\sqrt{2} - 2\sqrt{4-x}}{x-2}}\\\\ \quad = \displaystyle{2 \lim_{x \rightarrow 2} \frac{\sqrt{2} - \sqrt{4-x}}{x-2}}\\\\ \quad = \displaystyle{2 \lim_{x \rightarrow 2} \left[ \frac{\sqrt{2} - \sqrt{4-x}}{x-2} \cdot \frac{\sqrt{2} + \sqrt{4-x}}{\sqrt{2} + \sqrt{4-x}} \right]}\\\\ \quad = \displaystyle{2 \lim_{x \rightarrow 2} \frac{2-(4-x)}{(x-2)(\sqrt{2} + \sqrt{4-x})}}\\\\ \quad = \displaystyle{2 \lim_{x \rightarrow 2} \frac{(x-2)}{(x-2)(\sqrt{2} + \sqrt{4-x})}}\\\\ \quad = \displaystyle{2 \lim_{x \rightarrow 2} \frac{1}{(\sqrt{2} + \sqrt{4-x})}}\\\\ \quad = \displaystyle{2 \frac{1}{2\sqrt{2}}}\\\\ \quad = \displaystyle{\frac{1}{\sqrt{2}}} \end{array}$$From there, we proceed as we did with our first solution, finding the point $(2,1-2\sqrt{2})$ on the tangent line and using point-slope form to write the equation of the tangent line:
$$y - (1-2\sqrt{2}) = \frac{1}{\sqrt{2}} (x - 2)$$Find the velocity of a particle if the position is given by $\displaystyle{s(t) = \frac{2-t}{3t+1}}$ meters at $t$ seconds.
Recall that an instantaneous velocity can be considered a slope of a corresponding tangent line.
With this in mind, to find the velocity of a particle whose position is given by $s(t)$ at $t$ seconds, we just find $s'(t)$:
$$\begin{array}{rcl} s'(t) &=& \displaystyle{\lim_{h \rightarrow 0} \frac{s(t+h) - s(t)}{h}}\\\\ &=& \displaystyle{\lim_{h \rightarrow 0} \frac{\displaystyle{\frac{2-(t+h)}{3(t+h)+1} - \frac{2-t}{3t+1}}}{h}}\\\\ &=& \displaystyle{\lim_{h \rightarrow 0} \frac{(2-t-h)(3t+1) - (2-t)(3t+3h+1)}{h(3t+3h+1)(3t+1)}}\\\\ &=& \displaystyle{\lim_{h \rightarrow 0} \frac{-7h}{h(3t+3h+1)(3t+1)}}\\\\ &=& \displaystyle{\lim_{h \rightarrow 0} \frac{-7}{(3t+3h+1)(3t+1)}}\\\\ &=& \displaystyle{\frac{-7}{(3t+1)^2}} \end{array}$$Show, using limits, that the derivative of the function $(f(x))^2$ is $2 \cdot f(x) \cdot f'(x)$.
By definition, $\displaystyle{\frac{d}{dx} ((f(x))^2) = \lim_{h \rightarrow 0} \frac{(f(x+h))^2 - (f(x))^2}{h}}$.
We can factor the numerator as a difference of squares and use the limit laws to do the following:
$$\begin{array}{l} \displaystyle{\lim_{h \rightarrow 0} \frac{(f(x+h)+f(x))(f(x+h)-f(x))}{h}}\\\\ \quad = \displaystyle{\left[ \lim_{h \rightarrow 0} \, (f(x+h)+f(x)) \right]} \cdot \displaystyle{\left[ \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \right]}\\\\ \quad = 2f(x) f'(x) \end{array}$$If $f(x)$ is differentiable at $x=3$ and $f(3) = 7$, compute $\displaystyle{\lim_{x \rightarrow 3} f(x)}$. Make sure to justify your answer.
As $f(x)$ is differentiable at $x=3$, it must be continuous there.
As continuity at $x=3$ means $f(3)$ and $\displaystyle{\lim_{x \rightarrow 3} f(x)}$ must agree, the limiting value we seek must equal $f(3)=7$.
Compute $\displaystyle{\lim_{y \rightarrow 5} \frac{y^2}{2y+1}}$ using an appropriate limit law to justify each step.