Practice

In the previous lesson, we covered what a derivative is, and how to find it.

We learned that the standard formula to find the derivative of a function $f(x)$ is

$\lim_{\delta x \to 0}\frac{f(x+\delta x)-f(x)}{\delta x} = \frac{\delta y}{\delta x}.$

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$\displaystyle{\frac{d \sin{x}}{dx} = }$

Hint 1

Remember your $\sin{(a+b)}$ formula?

Hint 2

Try to use some of the identities we learned last chapter.

Solution 1
$\lim_{\delta x \to 0} \frac{\sin{(x+\delta x)} -\sin x}{\delta x}$
$\lim_{\delta x \to 0} \frac{\sin x\cos{\delta x}+\cos x\sin{\delta x} -\sin{x}}{\delta x}$

Distribute, and split the limit:

$\lim_{\delta x \to 0} (\sin x)\frac{\cos{\delta x} -1}{\delta x} + \lim_{\delta x \to 0}(\cos x)\frac{\sin x}{x}$

Look! There are our old friends, LI3 and LI4, hiding out in our practice problems! We can simply substitute, and move on:

$(\sin x)(0)+(\cos x)(1) = \boxed{\cos x}$

More neatly written,

$\frac{d \sin{x}}{dx} = \cos x.$

$\displaystyle{\frac{d\cos x}{dx}}=$

Hint 1

This question is solved almost the exact same way as the previous.

Solution 1
$\lim_{\delta x \to 0} \frac{\cos{(x+\delta x)} -\cos x}{\delta x}$
$\lim_{\delta x \to 0} \frac{\cos x \cos{\delta x} - \sin x \sin{\delta x} -\cos x}{\delta x}$

Remember that the limit operation is distributive across multiplication, division, addition, and subtraction. Below, we distribute amongst subtraction.

$\lim_{\delta x \to 0} \cos x (\frac{\cos{\delta x} -1}{\delta x}) - \lim_{\delta x \to 0} \sin x(\frac{\sin{\delta x}}{\delta x})$

The value of $x$ is independent of $\delta x$, the approaching variable, so we can remove values that are functions of $x$ to be a coefficient of the limit.

$\cos x \cdot\lim_{\delta x \to 0}\frac{\cos{\delta x} -1}{\delta x} - \sin x\cdot\lim_{\delta x \to 0}\frac{\sin{\delta x}}{\delta x}$

Aha! There's our LI3 and LI4! Substitute:

$\cos x \cdot 0 - \sin x\cdot 1 =-\sin x$

$\displaystyle{\frac{de^x}{dx}}=$

Hint 1

Try to use LI2.

Solution 1

Set up the limit:

$\lim_{\delta x \to 0} \frac{e^{x\delta x} - e^x}{\delta x}.$

Now, we can factor $e^x$ out of the limit:

$e^x\lim_{\delta x \to 0} \frac{e^{\delta x} - 1}{\delta x}.$

This is LI2! We can substitute LI2=1:

$\frac{de^x}{dx} = e^x.$

This is one of the most important differentiations, next to $\sin x$ and $\cos x$. This will be particularly useful in the coming chapters.

$\displaystyle{\frac{d \ln x}{dx}} =$

Hint 1

What was the formula for $\log_n{a} - \log_n{b}$?

Solution 1
$\lim_{\delta x \to 0} \frac{\ln{(x+\delta x)} -\ln x}{\delta x}$

Apply the difference of logarithms formula:

$\lim_{\delta x \to 0} \frac{1}{\delta x}\ln{(\frac{x+\delta x}{x})}$
$\lim_{\delta x \to 0} \frac{1}{\delta x}\ln{(1+\frac{\delta x}{x})}$

Now we apply the logarithm coefficient formula:

$\lim_{\delta x \to 0} \ln{[(1+\frac{\delta x}{x})^{\frac{1}{\delta x}}]}$
$\lim_{\delta x \to 0} \ln{[(1+\frac{\delta x}{x})^{\frac{x}{x\delta x}}]}$
$\lim_{\delta x \to 0} \ln{[(1+\frac{\delta x}{x})^{\frac{x}{\delta x}\cdot \frac{1}{x}}]}$
$\lim_{\delta x \to 0} \frac{1}{x}\ln{[(1+\frac{\delta x}{x})^{\frac{x}{\delta x}}]}$
$\frac{1}{x}\lim_{\delta x \to 0} \ln{[(1+\frac{\delta x}{x})^{\frac{x}{\delta x}}]}$

Wow! Doesn't it resemble LI5?

However, the actual approaching variable is $\delta x$, not $\frac{\delta x}{x}$. So, we can't immediately substitute for our identity.

So, I guess we're stuck now- there's nothing we can do. Or is there?

Let's step away from the problem and think about it for a moment. As $\delta x$ approaches 0, what does $\frac{\delta x}{x}$ approach?

We can just plug in 0 to the numerator, and we get 0! So now, we can say that

$\frac{1}{x}\lim_{\delta x \to 0} \ln{[(1+\frac{\delta x}{x})^{\frac{x}{\delta x}}]}=\frac{1}{x}\lim_{\frac{\delta x}{x} \to 0} \ln{[(1+\frac{\delta x}{x})^{\frac{x}{\delta x}}]}$

(Make sure you see why that step was necessary.)

Now, we can just substitute LI5 here:

$\frac{1}{x}\cdot \ln{e} = \boxed{\frac{1}{x}}.$

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