# Rookie derivative question

## Homework Statement

##f(x) = (5x+6)^{10} , f'(x)=?##

## Homework Equations

##\frac{d}{dx}x^n = nx^{n-1}##?
3. The Attempt at a Solution [/B]
I do know the solution ##f'(x) = 50(5x+6)^9##,but I don't know how this solution came to be.I downloaded this problem from the web and it only comes with the answer, not the process. First I thought that 50 comes from 10x5, but that was totally busted when I tested ##(4x+6)^{10}##, so I don't think this is power rule.

I also couldn't find things related to this in my textbook maybe because I don't know the keywords.

Sorry,I think I got it.Is that still the power rule, just treating the entire bottom thing like x?

Mentor

## Homework Statement

##f(x) = (5x+6)^{10} , f'(x)=?##

## Homework Equations

##\frac{d}{dx}x^n = nx^{n-1}##?
3. The Attempt at a Solution [/B]
I do know the solution ##f'(x) = 50(5x+6)^9##,but I don't know how this solution came to be.I downloaded this problem from the web and it only comes with the answer, not the process. First I thought that 50 comes from 10x5, but that was totally busted when I tested ##(4x+6)^{10}##, so I don't think this is power rule.

I also couldn't find things related to this in my textbook maybe because I don't know the keywords.
You need to use the chain rule form of the power rule. IOW, if ##f(x) = (u(x))^n##, then ##f'(x) = n(u(x))^{n - 1} \cdot u'(x)##.

• YoungPhysicist
You need to use the chain rule form of the power rule. IOW, if ##f(x) = (u(x))^n##, then ##f'(x) = n(u(x))^{n - 1} \cdot u'(x)##.
So in this case:

##\frac{d}{dx}(5x+6)^{10} = 10(5x+6)^9 \cdot 5 = 50(5x+6)^9##?

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• WWGD and fresh_42
Mentor
So in this case:

##\frac{d}{dx}(5x+6)^{10} = 10(5x+6)^9 \cdot 5 = 50(5x+6)^9##?
I do know the solution ##f'(x) = 50(5x+6)^9##,

Just making sure that the process is right.

Gold Member
Just to reinforce the concept, Young, can you figure out f'(x) for ##f(x)=e^{2x+6x^2} ##( e is Euler's constant)?

Mentor
Just making sure that the process is right.
But you can be pretty sure that your process is right if the answer you get is the answer that you already know. That should be a fairly strong clue that you're on the right track.

Homework Helper
Dearly Missed

## Homework Statement

##f(x) = (5x+6)^{10} , f'(x)=?##

## Homework Equations

##\frac{d}{dx}x^n = nx^{n-1}##?
3. The Attempt at a Solution [/B]
I do know the solution ##f'(x) = 50(5x+6)^9##,but I don't know how this solution came to be.I downloaded this problem from the web and it only comes with the answer, not the process. First I thought that 50 comes from 10x5, but that was totally busted when I tested ##(4x+6)^{10}##, so I don't think this is power rule.

I also couldn't find things related to this in my textbook maybe because I don't know the keywords.

If ##a## and ##b## are constants we have
$$\frac{d}{dx} \: (ax+b)^n = n (ax+b)^{n-1} \cdot \frac{d}{dx} (ax+b) = n a (ax+b)^{n-1}.$$
This is just a simple application of the chain rule.

• YoungPhysicist
archaic
Have a look at patrickjmt's youtube channel and website :
http://patrickjmt.com/ (choose the subject from the top)
In his website, the videos are ordered as follows : This is him explaining the chain rule :

Harder examples :

#### Attachments

• YoungPhysicist
• archaic
archaic
Nice vids.Thanks.
You're welcome.

Homework Helper
Gold Member

## Homework Statement

##f(x) = (5x+6)^{10} , f'(x)=?##

## Homework Equations

##\frac{d}{dx}x^n = nx^{n-1}##?
3. The Attempt at a Solution [/B]
I do know the solution ##f'(x) = 50(5x+6)^9##,but I don't know how this solution came to be.I downloaded this problem from the web and it only comes with the answer, not the process. First I thought that 50 comes from 10x5, but that was totally busted when I tested ##(4x+6)^{10}##, so I don't think this is power rule.

I also couldn't find things related to this in my textbook maybe because I don't know the keywords.

## Homework Statement

##f(x) = (5x+6)^{10} , f'(x)=?##

## Homework Equations

##\frac{d}{dx}x^n = nx^{n-1}##?p[/B]
.

I also couldn't find things related to this in my textbook maybe because I don't know the keywords.

It is probably either the very next thing in your textbook after how to differentiate xn. Or else even before.

Your book may or may not contain the wording "chain rule". I had never met it in my life before I did so on this forum. And for a time I had to look up every time I met it. What is the point of a name that conveys exactly nothing? When I was at school it came under "function of a function". Which conveys something. Symbolised f(g(x)). So learn what is the differentiation rule for that but do yourself another favour.

Instead of being dependent on learning forgettable rules take steps to make the damn thing obvious, forever!

Sketch or compute any function. Why not x10 ? Then try 5x10. Then (5x + 6)10. Try to predict how they will go, but in any case once you have got the curves see how they and their relations make sense, and make sense of these rules. See why they have to be.

Another thing you can do is sketch any old function, just any old curve, you don't have to have a formula for it, y = f(x) against x. Turn the paper through +90°, so the y-axis becomes horizontal, make the now vertical axis z, draw another different arbitrary curve which will this time be z = g(y). ( can if you like fold the paper along the y-axis and make a part vertical, not essential.) Now take a narrow strip of x which will be your dx as an idea. Follow through onto the rest of the diagram until you reach the points based from traced from x and (x + dx) on the g curve. (Alternative to curves just use straight line sections. )
By how much is the corresponding z going to increase? Depends on the slopes at the location of the strips.

Then all these common formulae for differeantials of sums, products, quotients, functions of functions, special functions like trig etc. can be and conventionally are obtained working from the general
$$\dfrac {df\left( x\right) }{dx}=\lim _{\delta \rightarrow 0}\dfrac {\left[ f\left( x+\delta x\right) -f\left( x\right) \right] }{\delta x }$$ but it is better to have a picture of what these derivations are saying.

By coming here you have helped yourself already, but by this active approach you will find you need to do so not so often.

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• 