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# Worked example: Derivative of √(3x²-x) using the chain rule

f(x)=√(3x²-x) is a composition of the functions √x and 3x²-x, and therefore we can differentiate it using the chain rule. Created by Sal Khan.

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• Does he ever explain why exactly we multiply the derivatives?

I'm having a hard time conceptualizing why this works and it's making implicit differentiation that much harder to understand since it's basically an extension of the chain rule. The whole fraction cancelling pseudo-explanation (d(*)*/d(*) * d*/dx = d(*)*/dx) unfortunately isn't good enough for my brain who's a stickler for understanding everything inside and out before accepting it. •   To understand chain rule think about definition of derivative as rate of change. d[f(g(x)]/d[x] basically means rate of change of f(g(x)) regarding rate of change of x, and to calculate this we need to know two values:
1- How much f(g(x)) changes while g(x) changes = d[f(g(x))]/d[g(x)]
2- How much g(x) changes while x changes = d[g(x)]/d[x]
to calculate rate of change of f(g(x)) in regard to rate of change of x, you just need to multiply these two values together because x changes f(x) and g(x) changes f(g(x)) (it should be obvious thinking about definition of a function in mathematics).

About fraction cancelling, as you mentioned it is a pseudo-explanation and speaking mathematics dg/df*df/dx is not like c/b*b/a = c/a so you can't cancel out nominator and denominator. (df/dx is not a division of df by dx)

Finally, here is a rigorous proof of the chain rule that might be helpful:
http://math.rice.edu/~cjd/chainrule.pdf
• I have a hard time understanding that the the square root of x is equal to x raised to 1/2. Can anyone show me a proof that square root of x=x^1/2? Also, if square roof of x=x^1/2, what would x^-1/2 be?

Thanks! •  Nice question !
A square root just gives a number that when multiplied with itself gives the number the square root is being found of. Like √4 = 2 As 4= 2*2
Let's take x^1/2 * x^1/2. The product rule of exponents is :
x^a * x^b = x^(a + b)
So by this x^1/2 * x^1/2 = x^( 1/2+1/2)= x^1 = x
Also now what is the square root of x ? As we saw above the square root is a number when multiplied with itself gives the number whose square root is being found. x^1/2 when multiplied by itself gives x like 2 gives 4. So √x= x^1/2
• For the example in the video: Is there really no way to simplify the final answer? • You could rewrite it as a fraction, (6x-1)/2(sqrt(3x^2-x)), but that's just an alternate form of the same thing rather than a true simplification. Sometimes the answer to a problem like this is messy. You should be prepared for messy answers when applying the product rule, the quotient rule and the chain rule.
• At , excactly why do we multiply with the derivative of g(x)? Is there any intuition here, or perhaps any proof one can watch? :-) • Unfortunately, I don't think that Khan Academy has a proof for chain rule.
I personally have not seen a proof of the chain rule. The reasoning that I use comes from the ideas function transformations.
We have the function f(x). When I do f(2x), that squeezes the graph in the horizontal direction by a factor of 2. My reasoning is that at a point, f(g(x)) will have the normal slope f'(g(x)), but it needs a multiplier because g(x) is instantaneously stretching or compressing in the x direction which will decrease or increase, respectively, the slope of the tangent line. This is accounted for by multiplying by g(x). That may or may not help you. I have found that I have some unusual ways of thinking about math.
Paul's Online Math notes happens to have a proof of the chain rule. I have not read it myself, but I put a link here so that you may enrich your learning if you so choose.
http://tutorial.math.lamar.edu/Classes/CalcI/DerivativeProofs.aspx#Extras_DerPf_ChainRule
• I'm having a hard time understanding where f(x) and g(x) come from. From -, how did you find that f(x) is the sqrt. of x and g(x) is 3x^2-x? • He takes them directly from the expression he is planning to differentiate. The square root is the outer function, 3x^2 - x is the inner function.

The x in the definition of f(x) is not the same as the x in the definition of g(x). They are independent functions that he combines into f(g(x)).

I guess you could say that f(g(x)) = √g(x) = √(3x^2 - x) and also
f(g(x)) = f(3x^2 - x) = √(3x^2 - x)
• I was trying to understand the difference(s) between `d/dx` and `f'(x)` through some vids but totally made me out of space. Can I rewrite `d/dx [f(g(x))] = f'(g(x)) * g'(x)` as `d/dx [f(g(x))] = d/d[g(x)] [f(g(x))] * d/dx [g(x)]`? Why do you have to use d/dx and f prime in a same math expression? • How do you prove that chain rule is always true?I mean,you can verify it by taking examples,but how do you prove it? • if i was given this question, how would i know straight away the difference between whats f(x) and g(x) by just looking at the square root of (3x^2 - x) ? at • This takes some practice with function composition. Often you can work your way from the outside in. Consider this quiz problem.
f(x) = sin^4(e^(x^2)) can be expressed as f(x) = a(b(c(d(x)))), find a, b, c, and d. Check your answer by recomposing the function according to your chosen a, b, c, and d.
a(x) = x^4
b(x) = sin(x)
c(x) = e^x (I almost wrote this as x^2 on the quiz)
d(x) = x^2 (And this one was almost e^x)
Recomposing the function saved me from mixing c and d.
c(d(x)) = e^(x^2))
b(c(d(x))) = sin(e^(x^2)))
a(b(c(d(x)))) = (sin(e^(x^2)))^4 or sin^4(e^(x^2))
This is a place where prerequisites become VERY important. The chain rule is a biggie, if you can't decompose functions it will trip you up all through calculus.
• When Sal applies the power rule to f(x), wouldn't it become 1/2*x^(-3/2) instead of 1/2*x^(-1/2) because the power rule does state that you subtract one from the exponent? • d/dx[sqrt(3x^2-x)]= 1/2[sqrt(3x^2-x)].............since d/dx[sqrt(x)] = 1/2(sqrtx)
m I missing something? isnt this right?
Plz explain. • You are missing the chain rule. Here is how you would find that derivative...
d/dx(sqrt(3x^2-x)) can be seen as d/dx(f(g(x))
where f(x) = sqrt(x) and g(x) = 3x^2-x
The chain-rule says that the derivative is: f'(g(x))*g'(x)
We already know f(x) and g(x); so we just need to figure out f'(x) and g'(x)
f"(x) = 1/sqrt(x) ; and ; g'(x) = 6x-1
Therefore d/dx(sqrt(3x^2-x)) = [1/sqrt(3x^2-x)*(6x-1)]
Which can be simplified to... (6x-1)/(sqrt(3x^2-x))

Hope that helps! :)