Calculus

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)f(x) is

limδx0f(x+δx)f(x)δx=δyδx.\lim_{\delta x \to 0}\frac{f(x+\delta x)-f(x)}{\delta x} = \frac{\delta y}{\delta x}.

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

Hint 1

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

Hint 2

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

Solution 1
limδx0sin(x+δx)sinxδx\lim_{\delta x \to 0} \frac{\sin{(x+\delta x)} -\sin x}{\delta x}
limδx0sinxcosδx+cosxsinδxsinxδx\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:

limδx0(sinx)cosδx1δx+limδx0(cosx)sinxx\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:

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

More neatly written,

dsinxdx=cosx.\frac{d \sin{x}}{dx} = \cos x.

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

Hint 1

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

Solution 1
limδx0cos(x+δx)cosxδx\lim_{\delta x \to 0} \frac{\cos{(x+\delta x)} -\cos x}{\delta x}
limδx0cosxcosδxsinxsinδxcosxδx\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.

limδx0cosx(cosδx1δx)limδx0sinx(sinδxδx)\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 xx is independent of δx\delta x, the approaching variable, so we can remove values that are functions of xx to be a coefficient of the limit.

cosxlimδx0cosδx1δxsinxlimδx0sinδxδx\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:

cosx0sinx1=sinx\cos x \cdot 0 - \sin x\cdot 1 =-\sin x

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

Hint 1

Try to use LI2.

Solution 1

Set up the limit:

limδx0exδxexδx.\lim_{\delta x \to 0} \frac{e^{x\delta x} - e^x}{\delta x}.

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

exlimδx0eδx1δx.e^x\lim_{\delta x \to 0} \frac{e^{\delta x} - 1}{\delta x}.

This is LI2! We can substitute LI2=1:

dexdx=ex.\frac{de^x}{dx} = e^x.

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

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

Hint 1

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

Solution 1
limδx0ln(x+δx)lnxδx\lim_{\delta x \to 0} \frac{\ln{(x+\delta x)} -\ln x}{\delta x}

Apply the difference of logarithms formula:

limδx01δxln(x+δxx)\lim_{\delta x \to 0} \frac{1}{\delta x}\ln{(\frac{x+\delta x}{x})}
limδx01δxln(1+δxx)\lim_{\delta x \to 0} \frac{1}{\delta x}\ln{(1+\frac{\delta x}{x})}

Now we apply the logarithm coefficient formula:

limδx0ln[(1+δxx)1δx]\lim_{\delta x \to 0} \ln{[(1+\frac{\delta x}{x})^{\frac{1}{\delta x}}]}
limδx0ln[(1+δxx)xxδx]\lim_{\delta x \to 0} \ln{[(1+\frac{\delta x}{x})^{\frac{x}{x\delta x}}]}
limδx0ln[(1+δxx)xδx1x]\lim_{\delta x \to 0} \ln{[(1+\frac{\delta x}{x})^{\frac{x}{\delta x}\cdot \frac{1}{x}}]}
limδx01xln[(1+δxx)xδx]\lim_{\delta x \to 0} \frac{1}{x}\ln{[(1+\frac{\delta x}{x})^{\frac{x}{\delta x}}]}
1xlimδx0ln[(1+δxx)xδx]\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 δx\delta x, not δxx\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 δx\delta x approaches 0, what does δxx\frac{\delta x}{x} approach?

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

1xlimδx0ln[(1+δxx)xδx]=1xlimδxx0ln[(1+δxx)xδx]\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:

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

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