Step 1: Composite Functions - One outer, one inner function
Derive with Chain Rule

Now, couple examples for you - identify inner/outer and find derivatives:

1) cos(10x-5)
2 ) sin(15x)
3) sin(x-9)

Answers are on the botoom of the page

Let's continue solving the examples:

3. $\sqrt{x^2+x+1}$

outer: $\sqrt()$ - square root as it is applied to the inner summation
inner: $x^2+x+1$

derivative of the outer: derivative of the square root $(\sqrt{x})'$ is $\frac{1}{2}x^{-\frac{1}{2}}$ BUT our argument is not just x but $x^2+x+1$ so the final derivative of the outer function is $\frac{1}{2}(x^2+x+1)^{-\frac{1}{2}}$

derivative of the inner: $(x^2+x+1)'=2x+1$

final derivative is the product of the inner and outer functions derivatives:

$(\sqrt{x^2+x+1})'=\frac{1}{2}\frac{2x+1}{(x^2+x+1)^{\frac{1}{2}}}$

Now let's focus only on determining inner and outer functions

4. $(5x+11)^2$

outer: $()^2$
inner: 5x+11
as square is applied to 5x+11

5. $(5x^2+11)^{\frac{3}{2}}$

outer: $()^\frac{3}{2}$
inner: $5x^2+11$
→ as power $\frac{3}{2}$ is applied to $5x^2+11$
or in another words $5x^2+11$ is inside (an argument) of $ ()^ \frac{3}{2}$

Now, couple examples for you:

4) $(x^3-2x)^2$

5 ) $\sqrt{x^5-1}$

Answers are at the botoom of the page

6. $\frac{1}{2}\sqrt{-x-3}$

what do you think?

outer: $\frac{1}{2}\sqrt()$ → $\frac{1}{2}$ times square root as it is applied to the inner function
inner: $-x-3$

7. $\frac{1}{4}ln(x^4)$

what do you think?

outer: $\frac{1}{4}ln() $ → as it is applied to $x^4$
inner: $x^4$

8. $e^{x^3+5}$

what do you think?:)

outer: $e^{()}$ → e it to the power of inner function
inner: $x^3+5$

9. $\frac{1}{5x}$

we can write $\frac{1}{5x}=(5x)^{-1}$

outer: $()^{-1}$ → raising to the power of minus one
inner: $5x$

10. $\frac{3}{4x}$

we can write $\frac{3}{4x}=3\cdot (4x)^{-1}$

what do you think?:)

outer: $3\cdot ()^{-1}$
inner: $4x$

11. $\frac{2}{3x^2}$

we can write $\frac{2}{3x^2}=2\cdot (3x^2)^{-1}$

what do you think?:)

outer: $2\cdot ()^{-1}$
inner: $3x^2$

OR
we can write equivalently:
$\frac{2}{3x^2}=2\cdot (3x^2)^{-1}=\mathbf{2\cdot (3x)^{-2}}$

outer: $2\cdot ()^{-2}$
inner: $3x$

Now, couple examples for you

Identify the inner and outer function:

6) $\frac{5}{(3x)^3}$

7) $\frac{2}{\sqrt {5x}}$

8) $e^{x^4+2}$

9) $2\cdot ln(2x-10)$

10) $e^{-x}$

Answers are at the bottom of the page

Now, you should be ready to calculate couple examples of derivatives of composite function. We will do 3 together and you will do 5 on your own.

Identify inner/outer function and find derivative

12. $3e^{5x^2-8}$

what do you think?

outer: $3e^{()}$ with derivative $3e^{5x^2-8}$ - keep the inner function as the argument!

inner: $5x^2-8$ with derivative $10x$

derivative:
($3\cdot e^{5x^2-8}$)'= $3e^{5x^2-8}\cdot 10x=30xe^{5x^2-8}$

now your turn:

11) $2e^{3x^5-x+1}$

12) $10e^{x^5-2x^4}$

Answers are at the bottom

13. $\frac{10}{(6x)^4}$

we can write equivalently:
$\frac{10}{(6x)^4}=10\cdot (6x)^{-4}$

outer: $10()^{-4}$ and its derivative: $-4\cdot 10()^{-5}$, plugging in the argument: $-4\cdot 10(6x)^{-5}$

inner: $ 6x$ with derivative 6

final derivative:

$\Big(\frac{10}{(6x)^4}\Big)$'= $-4\cdot 10(6x)^{-5}\cdot 6 $

14. $cos(sin(x))$

outer: $cos()$ and its derivative: $-sin()$, plugging in the argument in derivative: $-sin(cos(x))$

inner: $ sin(x)$ with derivative cos(x)

final derivative:

$(cos(sin(x)))'$=$ -sin(cos(x))\cdot cos(x) $

15. $ln(\frac{1}{x})$

outer: $ln()$ and its derivative: $\frac{1}{x}$,
plugging in the argument ($\mathbf{\frac{1}{x}}$) in derivative: $\frac{1}{\mathbf{\frac{1}{x}}}=x$

inner: $\frac{1}{x}=x^{-1} $, and its derivative: $-x^{-2}$

final derivative:

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

now your turn:

13) $\frac{11}{ln(x)}$

14) $sin(cos(x))$

15) $ln(sin(x))$


Check the Answers. Step 1 is completed.

Now, you should proceed to the Step 2 where you will practice harder examples with more that one inner function, like $ln(cos(x^2))$. Can you already tell what is the most inner, "less inner" and outer function?

Step 2: Two Inner Function Composition Chain Rule

Assuming you are confident in identifying the inner/outer function for simple composite functions like $tan(x^3)$, we can move to Step 2. Here, we will have two inner functions, like $tan(ln(5x))$.

Let's go! As usual, identify the inner/outer functions and differentiate.

We will start with an example. Let's consider a function:

$sin(ln(x^2))$

My preference is to always start from the most inner function. So, it must be the first thing we apply to x - here we square it.

⭢ So, the most inner function is $x^2$.

Then we take natural logarithm of it - ln()

⭢ So, the second inner function is natural logarithm $ln()$.

Then finally we apply outer function to it - sine:
⭢ outer function is sin()

Now the derivative of $sin(ln(x^2))$ is as you know from the Step 1 the product of derivatives of each inner and outer functions:

⭢ first inner function is $x^2$ with derivative 2x

⭢ second inner function is ln(argument) with derivative $\frac{1}{argument}$, in our case the argument is $x^2$, so we just plug it in: the derivative is $\frac {1}{x^2}$

⭢ outer function sin() with derivative cos() and plugging in the argument we get: $cos(ln(x^2))$

Final derivative is: $(sin(ln(x^2)))'=2x\cdot \frac{1}{x^2}\cdot cos(ln(x^2))$

Let's do 3 more examples together, and than you will do a couple on your own

Identify first inner, second inner and outer function and differentiate:

16. $e^{ln(x^2+1)}$

first inner: $x^2+1$

- and its derivative: 2x

second inner: ln(), its derivative is $\frac{1}{()}$ where in empty brackets is the argument of ln(): $x^2+1$. So $\frac{1}{x^2+1}$.

outer: $e^{()}$, its derivative is $e^{()}$ where in empty brackets is the argument of $e^{()}$: $ln(x^2+1)$. So $e^{ln(x^2+1)}$.

derivative of $e^{ln(x^2+1)}$ then is:

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

17. sin(cos(5x+2))

first inner: 5x+2
derivative is 5

second inner: cos()
its derivative is -sin()
plugging in the argument: $-sin(5x+2)$

outer: sin()
derivative is cos(), plugging in its argument: $cos(cos(5x+2))$

final derivative of this composite function:

$(sin(cos(5x+2)))'=\\=cos(cos(5x+2))(-sin(5x+2))\cdot 5$

18. $e^{cos^{2}(x)}$

first inner: cos(x) - because first cosine is applied to x and only after it is being squared
derivative is -sin(x)

second inner: $()^2$
with derivative: $2cos(x)$

outer: $e^{()}$
derivative is function itself $e^{()}$,
plugging in its argument: $e^{cos^{2}(x)}$

final derivative:

$(e^{cos^{2}(x)})'=e^{cos^{2}(x)} 2cos(x)(-sin(x)) $

Now, a couple similar examples for you - identify inner/outer and find derivatives:

16) $e^{cos(ln(x))}$
17) cos(sin(15x))
18) sin(sin(x+2))

Answers are at the botoom
of the page

Let's continue solving the examples:

19. $tan(\sqrt{x^2+x+1})$

first inner: $x^2+x+1$ with derivative 2x+1

second inner: $()^{\frac{1}{2}}$ with derivative $\frac{1}{2}()^{-\frac{1}{2}}$ plugging in the argument: $\frac{1}{2}(x^2+x+1)^{-\frac{1}{2}}$

outer: $tan()$ with derivative $sec ^{2}()$,
plugging in the argument: $sec ^{2}(\sqrt{x^2+x+1})$ - I will do it in one step from now

final answer:
$(tan(\sqrt{x^2+x+1}))'=\\=( 2x+1)\cdot \frac{1}{2}(x^2+x+1)^{-\frac{1}{2}}\times \\ \times sec ^{2}(\sqrt{x^2+x+1})$

Example for you: Identify and Derive
19) $ tan(cos^{2}(x))$

Let's do more together:

20. $(log_{8}(log_{10}(x)-5))^2$

first inner: $log_{10}(x)-5$
with derivative $\frac{1}{xln(10)}$ (have the table of derivatives with you:) )

second inner: $log_{10}()$
with derivative: $\frac{1}{(log_{10}(x)-5)ln(10)}$

outer $()^2$ as square is applied to everything that is inside
with derivative : $2(log_{10}(log_{10}(x)-5))$

final answer:
$ \Big((log_{10}(log_{10}(x)-5))^2\Big)'= \\= \frac{1}{xln(10)}\frac{1}{(log_{10}(x)-5)ln(10)} 2\times \\ \times (log_{10}(log_{10}(x)-5))$

Finally, couple examples for you and Step 2
is done

Identify the inner and outer function:

20) $log_{8}e^{2x}$

21) $log_{5}^2 (sin(x))$


Answers.

Practice identifying inner/outer function in the next worksheet in Step 3.