Question

DERIVATIVES
https://i.gyazo.com/eb8c7bb9208d6c0dfc977af786f24433.png

Answer 1

Yesterday you said tomorrow

Answer 2

Just do it!

Answer 3

Nothing is impossible!

Answer 4

Ok ill try to help now

Answer 5

oh no... this is torture. Ill be on my way now, didn't realize this was high school math... bye...

Answer 6

If you learned chain rule, and how to plug in numbers, you should be fine:)

Answer 7

we're doing product and quotient rule

Answer 8

So, you haven't ever done chain rule?

Answer 9

no

Answer 10

Oh, ok, we can go through it if you like:)

Answer 11

You know the derivative of \(x^n\).
\(\color{#000000}{ \displaystyle \frac{d}{dx} x^9=9x^{9-1}=9x^8 }\)

this is a basic power rule.

What if we had instead the following function,
\(\color{#000000}{ \displaystyle f(x)=(x^3+ 2x)^9 }\)
then, how would you differentiate this?

Expanding is certainly possible, but there is got to be a better way to differentiate this function (before my child is 35 years old).

\(\color{red}{**}\)
You are going to treat \(\color{#000000}{ \displaystyle (x^3+ 2x) }\) as just \(x\), when differentiating, and BY THE CHAIN RULE, you will multiply times the derivative of the inside.

So, in this case, you have:
\(\color{#000000}{ \displaystyle \frac{d}{dx}(x^3+ 2x)^9 = 9(x^3+ 2x)^8\color{blue}{\times\left[\frac{d}{dx}(x^3+ 2x)\right]} }\)

You know how to differentiate inside the blue.
The derivative of \(x^3\) is \(3x^2\) and the derivative
of \(2x\) is just \(2\). So you get:

\(\color{#000000}{ \displaystyle \frac{d}{dx}(x^3+ 2x)^9 = 9(x^3+ 2x)^8\color{blue}{\times\left[3x^2+ 2\right]} }\)

And you may write the final answer as:
\(\color{#000000}{ \displaystyle \frac{d}{dx}(x^3+ 2x)^9 = 9(3x^2+ 2)(x^3+ 2x)^8 }\)

Answer 12

it takes forever to type these equations, but you will see that differentiation process is ridiculously hard. In fact the opposite.

Answer 13

I will give you two more examples of a Chain rule, and I will then explain how this idea relates to this question.

Answer 14

Suppose I want to differentiate: \(\color{#000000}{ \displaystyle f(x)=\sin(x^5) }\).
Your previous tools like Product and Quotient Rule won't help.
The only way to do this ((asides writing a Taylor Series and term-by-term differentiating, or asides from using the first principles of differentiation)) ** is to apply the Chain Rule.

You will differentiate \(\color{#000000}{ \displaystyle \sin(x^5) }\) as if it were just a \(\color{#000000}{ \displaystyle \sin(x) }\), and then, you will multiply times the derivative of the inner function. (In this case, times the derivative of \(x^5\).)

So, ...
\(\color{#000000}{ \displaystyle \frac{d}{dx}\sin(x^5)=[\cos(x^5)]\color{blue}{\times \left[\frac{d}{dx} x^5\right]} }\)
The derivative of \(x^5\) is easy....
\(\color{#000000}{ \displaystyle \frac{d}{dx}\sin(x^5)=[\cos (x^5)]\color{blue}{\times \left[5x^4\right]} }\)

So, you can write the final answer as:
\(\color{#000000}{ \displaystyle \frac{d}{dx}\sin(x^5)=5x^4\cos (x^5) }\)

Answer 15

I will do the third example, and as you read, let me know if you understand this Chain Rule somewhat....

Answer 16

Now, I want to differentiate \(\color{#000000}{ \displaystyle f(x)=\sqrt{x^5+4e^x} }\).

Again, previous guys aren't of any help (unless you want to get these derivatives wrong and fail). So, the CHAIN RULE is applied.

Before that, I will re-write my function:
\(\color{#000000}{ \displaystyle f(x)=\left(x^5+4e^x\right)^{1/2} }\)

Now, I will differentiate using the chain rule.
\(\color{#000000}{ \displaystyle \frac{d}{dx}\left(x^5+4e^x\right)^{1/2} = \left(x^5+4e^x\right)^{~(1/2)~-1} \color{blue}{\times \left[\frac{d}{dx}\left(x^5+4e^x\right)\right]} }\)

The derivative of \(4e^x\) is same - it is \(4e^x\).
The derivative of \(x^5\) is just \(5x^4\), via the power rule.

\(\color{#000000}{ \displaystyle \frac{d}{dx}\left(x^5+4e^x\right)^{1/2} = \left(x^5+4e^x\right)^{~(1/2)~-1} \color{blue}{\times \left[5x^4+4e^x\right]} }\)

I will algebraically re-write the first (black) part, first:
\(\color{#000000}{ \displaystyle \frac{d}{dx}\left(x^5+4e^x\right)^{1/2} = \left(x^5+4e^x\right)^{-1/2} \color{blue}{\times \left[5x^4+4e^x\right]} }\)

\(\color{#000000}{ \displaystyle \frac{d}{dx}\left(x^5+4e^x\right)^{1/2} = \frac{1}{\left(x^5+4e^x\right)^{1/2} }\color{blue}{\times \left[5x^4+4e^x\right]} }\)

Next, I will multiply the blue part, and I will re-write the denominator as a square-root, since that's what it really is.

\(\color{#000000}{ \displaystyle \frac{d}{dx}\left(x^5+4e^x\right)^{1/2} = \frac{5x^4+4e^x}{\sqrt{x^5+4e^x} } }\)

And that is the final answer.

Answer 17

Let me know if the chain rule makes sense, as you read this.
(If you want more resources, or proves, I can link you or put them up myself)

Answer 18

You there?

Answer 19

From previous you should know that:
\(\color{#000000}{ [1]\quad \quad \displaystyle \frac{d}{dx}f(x)=f'(x) }\)

In other words, the derivative of the function is it's derivative.
---------------------------------
Applying the Chain Rule you should also
know that while it is TRUE to say that:
\(\color{#000000}{ \displaystyle \frac{d}{dx}G(x)=G'(x) }\)

it would be FALSE to claim that
\(\color{#000000}{ [2]\quad \quad\displaystyle \frac{d}{dx}G[f(x)]=G'[f(x)] }\)

Rather, the correct way to differentiate in such a case is:
\(\color{#000000}{ \displaystyle \frac{d}{dx}G[f(x)]=G'[f(x)] \cdot f'(x) }\)

Just like we had always been doing, we apply the Chain Rule.
THAT IS, we multiply times the derivative of the inner part.
(And so we have done for \(\color{#000000}{ \displaystyle \frac{d}{dx}G[f(x)]}\) )

Answer 20

I am typing an example similar to your problem.

Answer 21

--------------------------------------------
Example Problem Similar to yours.

Find \(\color{#000000}{ \displaystyle F'(0) }\), given that:
\(\color{#000000}{ \displaystyle f(0)=4 }\), \(\color{#000000}{ \displaystyle g(0)=-5 }\), \(\color{#000000}{ \displaystyle f'(0)=-6 }\), \(\color{#000000}{ \displaystyle g'(0)=7 }\)

Where the function \(\color{#000000}{ \displaystyle F(x) }\) is:
(A) \(\color{#000000}{ \displaystyle F(x)=5f(x)+4g(x) }\)
(B) \(\color{#000000}{ \displaystyle F(x)=f(x)g(x) }\)
(C) \(\color{#000000}{ \displaystyle F(x)=\left[g(x)\right]^3 }\)
(D) \(\color{#000000}{ \displaystyle F(x)=g(x)/f(x) }\)
--------------------------------------------
SOLUTION:


\(\color{red}{■}\) Part A)
You know that the function is:
\(\color{#000000}{ \displaystyle F(x)=5f(x)+4g(x) }\)

To differentiate this function you just differentiate
term-by-term and keep in mind rule [1] that I posted.
So, you get the following derivative:
\(\color{#000000}{ \displaystyle F'(x)=5f'(x)+4g'(x) }\)

Since you are to find \(\color{#000000}{ \displaystyle F'(0) }\), plug in \(\color{#000000}{ \displaystyle x=0 }\). This gives,
\(\color{#000000}{ \displaystyle F'(0)=5f'(0)+4g'(0) }\)

Since you know that: \(\color{#000000}{ \displaystyle f'(0)=-6 }\), \(\color{#000000}{ \displaystyle g'(0)=7 }\), therefore,
\(\color{#000000}{ \displaystyle F'(0)=5\cdot (-6)+4\cdot 7=-30+28=\color{red}{-2 } }\) <= Answer


\(\color{red}{■}\) Part B)
You know that the function is:
\(\color{#000000}{ \displaystyle F(x)=f(x)g(x) }\)

To differentiate this function you use the product rule,
ad again keep in mind rule [1].
\(\color{#000000}{ \displaystyle F'(x)=f'(x)g(x)+f(x)g'(x) }\)

Again, since you are to find \(\color{#000000}{ \displaystyle F'(0) }\), plug in \(\color{#000000}{ \displaystyle x=0 }\).
\(\color{#000000}{ \displaystyle F'(0)=f'(0)g(0)+f(0)g'(0) }\)

You know that:
\(\color{#000000}{ \displaystyle f(0)=4 }\), \(\color{#000000}{ \displaystyle g(0)=-5 }\)
\(\color{#000000}{ \displaystyle f'(0)=-6 }\), \(\color{#000000}{ \displaystyle g'(0)=7 }\).

Plug this, and you get:
\(\color{#000000}{ \displaystyle F'(0)=(-6)\cdot(-5)+(4)\cdot(7) }\)
\(\color{#000000}{ \displaystyle F'(0)=30+28=\color{red}{58} }\) <= Answer


\(\color{red}{■}\) Part C)
You know that the function is:
\(\color{#000000}{ \displaystyle F(x)=\left[g(x)\right]^3 }\)

Now, you need the chain rule to differentiate this!
\(\color{#000000}{ \displaystyle F'(x)=3\left[g(x)\right]^2\cdot g'(x) }\)

(See? I treated g(x) like x, and then multiplied times the derivative of the inside, or times the derivative of g(x), which is g'(x).)

And again plug \(\color{#000000}{ \displaystyle x=0 }\), since you want the \(\color{#000000}{ \displaystyle F'(0) }\).
\(\color{#000000}{ \displaystyle F'(0)=3\left[g(0)\right]^2\cdot g'(0) }\)

Now, I know that:
\(\color{#000000}{ \displaystyle g(0)=-5 }\), \(\color{#000000}{ \displaystyle g'(0)=7 }\).

\(\color{#000000}{ \displaystyle F'(0)=3\left[g(0)\right]^2\cdot g'(0) }\)
\(\color{#000000}{ \displaystyle F'(0)=3\left[-5\right]^2\cdot 7 }\)
\(\color{#000000}{ \displaystyle F'(0)=3\cdot 25\cdot 7 }\)
\(\color{#000000}{ \displaystyle F'(0)=\color{red}{525} }\) <= Answer


\(\color{red}{■}\) Part D)
Your function is:
\(\color{#000000}{ \displaystyle F(x)=g(x)/f(x) }\)

Apply the Quotient Rule, (which you can always prove using logarithmic differentiation if you have ever done that before, or when you do it in the future.)

\(\color{#000000}{ \displaystyle F'(x)=\frac{g(x)f'(x)-g'(x)f(x)}{[f(x)]^2} }\)

Again, you want \(\color{#000000}{ \displaystyle F'(0) }\), so plug \(\color{#000000}{ \displaystyle x=0 }\).

\(\color{#000000}{ \displaystyle F'(0)=\frac{g(0)f'(0)-g'(0)f(0)}{[f(0)]^2} }\)

You know that:
\(\color{#000000}{ \displaystyle f(0)=4 }\), \(\color{#000000}{ \displaystyle g(0)=-5 }\)
\(\color{#000000}{ \displaystyle f'(0)=-6 }\), \(\color{#000000}{ \displaystyle g'(0)=7 }\).

\(\color{#000000}{ \displaystyle F'(0)=\frac{g(0)f'(0)-g'(0)f(0)}{[f(0)]^2}=\frac{(-5)\cdot (-6)-(7)\cdot (4)}{[4]^2} }\)

\(\color{#000000}{ \displaystyle F'(0) =\frac{30-28}{16}=\color{red}{\frac{1}{8}}}\) <= Answer

Answer 22

Good Luck!