Simple Practice Problems

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

    Hint 1 Remember your $\sin{(a+b)}$ formula?
    Hint 2 Try to use some of the identities we learned last chapter.
    Show/Hide Solution $$\lim_{\delta x \to 0} \frac{\sin{(x+\delta x)} -\sin x}{\delta x}$$ $$\lim_{\delta x \to 0} \frac{\sin x\cos{\delta x}+\cos x\sin{\delta x} -\sin{x}}{\delta x}$$

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

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

    Hint This question is solved almost the exact same way as the previous.
    Show/Hide Solution $$\lim_{\delta x \to 0} \frac{\cos{(x+\delta x)} -\cos x}{\delta x}$$ $$\lim_{\delta x \to 0} \frac{\cos x \cos{\delta x} - \sin x \sin{\delta x} -\cos x}{\delta x}$$

    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:

    $$\cos x \cdot 0 - \sin x\cdot 1 =-\sin x$$
  3. $\displaystyle{\frac{de^x}{dx}}= $

    HintTry to use LI2.
    Show/Hide Solution

    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.

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

    Hint What was the formula for $\log_n{a} - \log_n{b}$?
    Show/Hide Solution $$\lim_{\delta x \to 0} \frac{\ln{(x+\delta x)} -\ln x}{\delta x}$$

    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:

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