Question

Super-uper urgent basic "differentiate with respect to x" problem: sqrt of [3 - 4x]

Since my booklet says formula is x^n = n*x^(n-1)

I got

0.5*(3-4x)^(-1/2)

But my teacher wrote in my failed exam that correct answer is:

0.5*(3-4x)^(-1/2) , (-4)

I don't get it at all why is this so

Answer 1

You did that part right, but you need to apply the chain rule as well.
You want \[f^{\prime}(x)= \frac{df}{du}\frac{du}{dx}\]

Answer 2

\[u=3-4x\]
so \[f(u)=(u)^{1/2}\]
and \[f^{\prime}(u) = \frac{1}{2}u^{-1/2}\]
But you are instructed to find f(x).

Answer 3

\[\frac{du}{dx}=-4\]
so
\[f^{\prime} = \frac{1}{2}u^{-1/2}(-4)\] which becomes
\[f^{\prime}(x) =-2(3-4x)^{-1/2}\] when you back substitute what u was

Answer 4

okay so after ive done the process of finding derivative i need to use chain's rule after that. im still trying to figure out how it works with your explanation hang on

Answer 5

du/dx is the coefficient of the variable we are differentiating with respect to, in this case x?

Answer 6

what is df/du

Answer 7

yes

Answer 8

with the substitution
\[f(u)=u^{1/2}\] so df/du is
\[\frac{1}{2}u^{-1/2}\]

Answer 9

df/du is the thing i found and thought was the final answer? so i get df/du by differentiating the specified thing with a variable and multiply this by the coefficient of the variable we are differentiating with respect to?

Answer 10

specified thing with variable is sqrt of 3-4x

Answer 11

not the coefficient the derivative of u with respect to the original variable you are differentiating with

Answer 12

So if F(x) = (stuff)^5 then the derivative would be
5(stuff)^(5-1) times the derivative of stuff with respect to x

Answer 13

okay the derivative is gotten by using the specified formula for a type of case. in addition to this (df/du) i will multiply this by du/dx - which is just coefficient of x? correct me if/how im wrong with the last one if i am

Answer 14

Example \[g(x) = (4x)^3\]
\[g^{\prime}(x) = 3(4x)^2\cdot \frac{d}{dx}(4x)\]

Answer 15

d/dx (4x) you just wrote there becomes just 4?

Answer 16

Yeah in this case. because the derivative of ax is a.
If instead you had something weird like
\[1-x^7\] in the paretheses you would have multiplied by \[-7x^6\]

Answer 17

yes

Answer 18

Another example
\[j(x) = (2x^2 +x)^4\] then the derivative would be
\[j^{\prime}(x) = 4(2x^2+x)^3\cdot (4x+1)\]
since \[\frac{d}{dx}(2x^2+x)=(4x+1)\]

Answer 19

ah so in a bit more difficult case you have two variables or more inside the brackets. you just differentiate what is inside the brackets by the normal rules so that you get 2x^2 -> 4x and x -> 1 and this is what is now what was 4 in my exercise!

Answer 20

any time when differentiating with respect to [variable] i have to do both take appropriate derivatives and then multiply them together for the final answer - except when i have e^x ?

Answer 21

\[f(x) = e^{2x+5}\]
u=2x+5
so
\[f^{\prime}(x) = e^{2x+5}(\frac{d}{dx}(2x+5))\]

Answer 22

So
\[f^{\prime}=2e^{2x+5}\]
The chain rule is used for when you substitute something more complicated for x.

Answer 23

more complicated for x - so you mean in this case the more complicated for x is 2x+5 ?

Answer 24

yes.

Answer 25

so in what you did i have to do something extra hmm let's see - so i take the exponent of the e^x and take the derivate of that. that results in 2+0. the 2 is multiplied by the derivative of e^x as a whole, which is just e^x (so e^2x+5 is e^2x+5)

Answer 26

\[=1/2(3-4x)^-1/2(-4)\]

Answer 27

Yes, exactly. This way they teach you something simpler like the derivative wrt x and then you learn how to do more complicated stuff.

Answer 28

nx^(n-1)d/dx(base)

Answer 29

as a fist rule i always first use formula to get dx/dy and if there is a composite (i think this is what it is, for example exponent with more than just x terms) i differentiate this composite to be multiplied with the derivative of dx/dy - if i understood right

Answer 30

yes, but try to use the substitution of variables.
You need to get used to substituting variables left and right and not getting confused. It takes a while.

Answer 31

so the derivative of composite takes the role of du/dx

Answer 32

what do you mean exactly by substitution of variables?

Answer 33

yes. or just some other variable other than x if they doing df/dx

Answer 34

see my reply you wiil get ypur problem

Answer 35

thanks 0322 i got it but freewilly had already helped the same before :)

Answer 36

Substitution of variables takes the difficult problem and writes it into something you recognize. You might not see
\[ e^{5x^2 \cos(x)}\] and know how to deal with it, but you know how to deal with \[e^u\]
because \[\frac{d}{du} e^u = e^u\]

Answer 37

That way when you are given a monster of a problem instead of giving up you can break it into smaller doable pieces.

Answer 38

yes sounds about right. in that example i must identify that the derivative dx/dy of the first of the two multiples (e^stuff being the first and cos(x) being the second) is just the expression e^(5x^2) itself. therefore i multiply that by what comes out of derivating (5x^2), that is, 10x

Answer 39

since that is 10x and we have the e^(5x^2) half of the problem is solved, so 10x*1^(5x^2)

Answer 40

then as we have broken this into two there's just cos(x) on the right side that becomes -sinx
so the f inal answer is -sinx multiplied by the 10x* thing i wrote

Answer 41

i usually do well when people explain this stuff to me with examples but in the exams that i have i tend to panic because of the variations of problems that there are :( i would have more questions if you had time about my exam area of mainly basic calculus

Answer 42

thanks so much so far because this is as good or better than my good math teacher's explanations of the derivative and differentiation and chain's rule as a whole

Answer 43

The derivative of \[5x^2\cos(x)\] wasn't something I wanted you to do. It was more of something that wasn't trivial like 4x.
you need the product rule for this one
\[e^{5x^2\cos{x}}(10x\cos(x)-5x^2\sin(x))\]

Answer 44

ooh i almost forgot about this

Answer 45

The best way is to understand the different rules with simple memorable examples that you can then see the rules when you are freezing up and apply them to the harder problem. I have to go, but good luck!

Answer 46

is e^(5x^2)*cosx really what u wanted to type there?

Answer 47

if u answer that then id be more than happy to let you go : D