Question

c(d) = 3√(5d^2 − 1)^5

Answer 1

\[\sqrt[3]{(5d^{2}-1)^{5}}\]
Find the derivative in the simplest factored form

Answer 2

Oh derivatives ... nice

Answer 3

Ok, what woud be the derivative of \(\large\color{#000000 }{ \displaystyle \sqrt[3]{x} }\) ?

Answer 4

HINT: \(\large\color{#000000 }{ \displaystyle \sqrt[3]{x} =x^{1/3} }\)

Answer 5

\[x^{-2/3}\]

Answer 6

you forgot something

Answer 7

The power rule:

\(\large\color{#000000 }{ \displaystyle \frac{d}{dx} x^n=\color{red}{n} x^{n-1}}\)

Answer 8

you woud have to multiply times 1/3

Answer 9

oh yeah

Answer 10

The power rule:

\(\large\color{#000000 }{ \displaystyle \frac{d}{dx} x^{1/3}=\color{red}{(1/3) } x^{-2/3}}\)

Answer 11

Woud you mind if we use, just:

\(\large\color{#000000 }{ \displaystyle f(x)=\sqrt[3]{(5x^{2}-1)^{5}} }\)

because to write derivatives as d/dd is a bit wierd
??

Answer 12

lets use x instead if you don;t mind

Answer 13

okay

Answer 14

i think i figured it out

Answer 15

now do the chain rue

Answer 16

chain rule**

Answer 17

\(\large\color{#000000 }{ \displaystyle f(x)=\sqrt[3]{(5x^{2}-1)^{5}} }\)


You know that,

\(\large\color{#000000 }{ \displaystyle \frac{d}{dx} \sqrt[3]{x}=\frac{1}{3}x^{-2/3}}\)

But inside the cube root you have another function of x.
(Right?)

So you need to apply the chain rule, as follows:

\(\large\color{#000000 }{ \displaystyle f(x)=\sqrt[3]{(5x^{2}-1)^{5}} }\)

\(\large\color{#000000 }{ \displaystyle f'(x)=\frac{1}{3} \left[(5x^{2}-1)^{5}\right]^{-2/3} \color{blue}{\times \frac{d}{dx} (5x^{2}-1)^{5}} }\)

Answer 18

have you seen the chain rue previously?

Answer 19

yeah

Answer 20

So, do you understand everything we have done so far?

Answer 21

yup , I think i figured it out

Answer 22

Ok, now you need to differentiate
\(\large\color{#000000 }{ \displaystyle \color{blue}{ (5x^{2}-1)^{5}} }\)

Answer 23

((You will need anothe Chain Rule))

Answer 24

What would be the derivative of x^5 ?

Answer 25

5x^4

Answer 26

Yes, but since instead of:
\(\large\color{#000000 }{ \displaystyle \color{blue}{ (\color{red}{x})^{5}} }\)

You have,
\(\large\color{#000000 }{ \displaystyle \color{blue}{ (\color{red}{5x^{2}-1})^{5}} }\)

You therefore must apply a chain rule (the second time)

Answer 27

What is the derivative of 5x^2-1 ?

Answer 28

10x

Answer 29

yes very good.

Answer 30

Now I will post the entire work for the derivative..
I wil try to be as clear and brief as I can

Answer 31

\(\color{#000000 }{ \displaystyle f(x)=\sqrt[3]{(5x^{2}-1)^{5}} }\)


we know that, \(\color{#000000 }{ \displaystyle \frac{d}{dx} \sqrt[3]{x}=\frac{1}{3}x^{-2/3}}\)

But inside the cube root you have another function of x.
So we apply the Chain Rule (in blue), as follows:

\(\color{#000000 }{ \displaystyle f'(x)=\frac{1}{3} \left[(5x^{2}-1)^{5}\right]^{-2/3} \color{blue}{\times \frac{d}{dx} (5x^{2}-1)^{5}} }\)


we know that, \(\color{#000000 }{ \displaystyle \frac{d}{dx} x^5=5x^4 }\)

But we have a function of x raised to the 5th power (not just x),
and therefore the expression inside the 5th power needs a Chain Rule (in green).

\(\color{#000000 }{ \displaystyle f'(x)=\frac{1}{3} \left[(5x^{2}-1)^{5}\right]^{-2/3} \color{blue}{\times 5(5x^{2}-1)^{4}} \color{green}{\times \frac{d}{dx} \left(5x^2 -1\right)}}\)


We know that: \(\color{#000000 }{ \displaystyle \frac{d}{dx} (5x^2-1)=2\cdot5x+0= 10x }\)

So we can write the derivative as:
\(\color{#000000 }{ \displaystyle f'(x)=\frac{1}{3} \left[(5x^{2}-1)^{5}\right]^{-2/3} \color{blue}{\times 5(5x^{2}-1)^{4}} \color{green}{\times 10x}}\)

Answer 32

Now all is needed is to simplify using algebra.

Answer 33

I wil post properties that might be helpful ...

Answer 34

\(\color{#bb4400}{ {\rm [1]}\quad\quad \displaystyle {\bf y}^{-A}= \frac{1}{{\bf y}^{A}} \quad\quad\Longrightarrow \quad\quad \displaystyle {\bf y}^{-N/M}= \frac{1}{{\bf y}^{N/M}} }\)

\(\color{#bb4400}{ {\rm [2]}\quad\quad \displaystyle {\bf y}^{Q} \times{\bf y}^{R} ={\bf y}^{Q+R} }\)