Advanced derivative examples

A few examples of derivatives that are harder than the basic work with formulae. If you need the definition of the derivative explained, go to the article on the derivative of a function.

First example

We start with something that is not too hard yet.

$$f(x)=x^3\cdot e^{-x}$$

From the derivative formulae we know that the derivative of e^x is again e^x. Unfortunately we cannot use this simple rule directly in this example, because the exponent is not just x but −x, so we have to treat the function as a composite function. In the first step, though, we still start by splitting it up using the product formula.

$$f^{\prime }(x)=(x^3{)}^{\prime }\cdot e^{-x}+x^3\cdot (e^{-x}{)}^{\prime }=$$

We differentiate the polynomial on the left using the standard formula and leave the rest the same:

$$=3x^2\cdot e^{-x}+x^3\cdot (e^{-x}{)}^{\prime }=$$

Now we have to differentiate the last function. As we said, it is a composite function, so we have to differentiate it using the rule for composite functions:

$$\begin{array}{rcl}f(x)& =& h(g(x))\\ f^{\prime }(x)& =& h^{\prime }(g(x))\cdot g^{\prime }(x)\end{array}$$

So the derivative of our function looks like this:

$$(e^{-x}{)}^{\prime }=e^{-x}\cdot (-x{)}^{\prime }=e^{-x}\cdot (-1)=-e^{-x}$$

The expression e^−x stays the same, because the derivative of e^x is again e^x, and in the first part of the formula we differentiate the outer function and leave the inner function undifferentiated. In the second part we multiply our intermediate result by the derivative of the input of the function, which is the function −x. The derivative of −x is −1. Now we just substitute our result into the previous expression:

$$=3x^2\cdot e^{-x}+x^3\cdot (e^{-x}{)}^{\prime }=3x^2\cdot e^{-x}-x^3\cdot e^{-x}.$$

Second example

The second example:

$$f(x)=\frac{\ln (\cos x)}{\tan x}$$

Here we have a fraction, with one function in the numerator and another in the denominator. So in the first step we differentiate using the quotient formula. After applying the formula we get:

$$\frac{(\ln (\cos x){)}^{\prime }\cdot \tan x-\ln (\cos x)\cdot (\tan x{)}^{\prime }}{{\tan }^{2}x}$$

We will work from the end and differentiate the tangent first, because we can do that simply using the basic formula. We have

$$(\tan x{)}^{\prime }=\frac{1}{{\cos }^{2}x}.$$

After applying this formula in the numerator of the fraction we get

$$\frac{(\ln (\cos x){)}^{\prime }\cdot \tan x-\ln (\cos x)\cdot \frac{1}{{\cos }^{2}x}}{{\tan }^{2}x}$$

Now we need to differentiate the logarithm. Unfortunately there is another function nested inside the logarithm, so we have a composite function and must differentiate the logarithm the harder way, as a composite function. The derivative of the logarithm on its own is 1/x, but here we substitute the input of the logarithm for x, that is, the cosine. Then we also have to multiply by the derivative of the input, that is, the derivative of the cosine. So the derivative of the logarithm looks like this:

$$(\ln (\cos x){)}^{\prime }=\frac{1}{\cos x}\cdot (\cos x{)}^{\prime }=-\frac{1}{\cos x}\cdot \sin x=-\frac{\sin x}{\cos x}$$

We got quite a nice fraction. If you remember your trigonometry well, you know that this fraction is equal to the tangent. So we can write that the whole thing is equal to minus the tangent:

$$(\ln (\cos x){)}^{\prime }=-\tan x$$

We substitute this into the previous result:

$$=\frac{(\ln (\cos x){)}^{\prime }\cdot \tan x-\ln (\cos x)\cdot \frac{1}{{\cos }^{2}x}}{{\tan }^{2}x}=\frac{-\tan x\cdot \tan x-\ln (\cos x)\cdot \frac{1}{{\cos }^{2}x}}{{\tan }^{2}x}=$$

In the numerator we have the tangent twice, so we can simply write it squared. The logarithm can be moved into the numerator of the following fraction.

$$=\frac{-{\tan }^{2}x-\frac{\ln (\cos x)}{{\cos }^{2}x}}{{\tan }^{2}x}=$$

Now we can split the numerator and get two fractions

$$=-\frac{{\tan }^{2}x}{{\tan }^{2}x}-\frac{\frac{\ln (\cos x)}{{\cos }^{2}x}}{{\tan }^{2}x}=$$

In the first fraction the numerator is equal to the denominator, so we get one. In the second fraction we can get rid of the fraction in the numerator by multiplying the top and bottom by cos²x.

$$=-1-\frac{\ln (\cos x)}{{\cos }^{2}x\cdot {\tan }^{2}x}=$$

Now we can write the tangent as the quotient sin(x)/cos(x). Because the tangent is squared, the numerator and denominator of that fraction are squared too. Then we can cancel cos²x straight away.

$$=-1-\frac{\ln (\cos x)}{{\cos }^{2}x\frac{{\sin }^{2}x}{{\cos }^{2}x}}=-\frac{\ln (\cos x)}{{\sin }^{2}x}-1.$$

That is the final result of the derivative.

Third example

Differentiate the function

$$f(x)=x^{(x^{\frac{1}{2}})}.$$

In case it is not clear – these are all powers, that is, “x to the power x to the power one half”. In the first step we rewrite the function using the identity:

$$x=e^{\ln x}$$

Using this identity we rewrite the first x in our example:

$$f(x)=(e^{\ln x}{)}^{x^{\frac{1}{2}}}$$

Now we apply the rule for working with powers:

$$(a^b{)}^c=a^{b\cdot c}$$

We remove the brackets and multiply the top exponent by the logarithm:

$$f(x)=e^{\ln x\cdot x^{\frac{1}{2}}}$$

Now we have the function in a form that is easier to differentiate. It is a composite function, and we will differentiate it as one. The first function is e^x, where we substitute the whole exponent for x, and the second function is in the exponent. Splitting it up, we get:

$$f^{\prime }(x)=e^{\ln x\cdot x^{\frac{1}{2}}}\cdot (\ln x\cdot x^{\frac{1}{2}}{)}^{\prime }$$

We will differentiate the bracket as a product, so we apply the product formula.

$$(\ln x\cdot x^{\frac{1}{2}}{)}^{\prime }=(\ln x{)}^{\prime }\cdot x^{\frac{1}{2}}+\ln x\cdot (x^{\frac{1}{2}}{)}^{\prime }=$$

The derivative of the logarithm is 1/x:

$$=\frac{1}{x}\cdot x^{\frac{1}{2}}+\ln x\cdot (x^{\frac{1}{2}}{)}^{\prime }=$$

Now we differentiate the second expression using the standard formula for differentiating power functions. So we have:

$${\left(x^{\frac{1}{2}}\right)}^{\prime }=\frac{1}{2}{x}^{\frac{1}{2}-1}=\frac{1}{2}{x}^{-\frac{1}{2}}=\frac{x^{-\frac{1}{2}}}{2}=\frac{1}{2x^{\frac{1}{2}}}=\frac{1}{2\sqrt{x}}$$

We substitute this result into our working:

$$=\frac{1}{x}\cdot x^{\frac{1}{2}}+\ln x\cdot \frac{1}{2\sqrt{x}}=$$

We can also simplify the first product, again using the rules for powers:

$$\frac{1}{x}\cdot x^{\frac{1}{2}}=x^{-1}\cdot x^{\frac{1}{2}}=x^{-1+\frac{1}{2}}=x^{-\frac{1}{2}}=\frac{1}{\sqrt{x}}$$

We rewrite the previous expression using the calculation we have just done:

$$=\frac{1}{\sqrt{x}}+\ln x\cdot \frac{1}{2\sqrt{x}}=$$

And finally we move the logarithm into the numerator of the fraction:

$$=\frac{1}{\sqrt{x}}+\frac{\ln x}{2\sqrt{x}}=$$

We can multiply the top and bottom of the first fraction by two, and then the fractions can be added:

$$=\frac{2}{2\sqrt{x}}+\frac{\ln x}{2\sqrt{x}}=\frac{2+\ln x}{2\sqrt{x}}$$

Now we can return to the whole derivative:

$$f^{\prime }(x)=e^{\ln x\cdot x^{\frac{1}{2}}}\cdot \left(\frac{2+\ln x}{2\sqrt{x}}\right)=$$

We replace e^{ln x} back with x:

$$=x^{\sqrt{x}}\cdot \left(\frac{2+\ln x}{2\sqrt{x}}\right).$$

That gives us the final result.