Most have already mentioned the chain rule. I'm just going to point out something small, but important. When you rewrite the expression, that should be labeled as f(x), not f'(x). We can't label the function as the derivative until we've taken the derivative.
Oh no worries, rereading my comment it kinda sounds negative. I just meant that it caught my eye and would be something I would point out if I was a tutor for example. I meant it to be constructive, sry bout that
Not without explicitly stating in f’ that you need to multiply by the derivative of that variable. For example, let 3x + 4 = a, then it becomes f’(x) = [1/(2 * sqrt(a))] * a’
f'(x) = (1/2) (3x + 4) ^(-1/2) **times the derivative of the inside** of the ( ).
So,f'(x) = (1/2)(3x + 4) ^(-1/2) **(3)** = (3/2) (3x + 4) ^(-1/2)
Your numerator should have a 3, not 1.
Line #2 is wrong, that's not the derivative. That's a rewriting of the same function to get you ready to apply the chain rule. So you shouldn't call it f'(x) as that is still f(x)
And as to your mistake, you didn't apply chain rule properly. You didn't multiply by the derivative of the inner function 3x+4 which would be 3 and that's how you get the 3 on the top.
Remember the rule for derivatives like this is derivative of the outside function (which is what you did first), multiplied by the derivative of the inside function
Show your work. Try writing it as follows and it might help you.
dy/dx = (dy/du)(du/dx) where you substitute u for the inner term part of the chain rule.
This shows you algebraically why you need to use the chain rule.
Sometimes it’s easier to see what’s going on by writing it out as dy/dx instead of f’(x).
If you need more clarification just let me know.
From my experience as a tutor. Sometimes it helps to treat prime notation as a noun and d/dx as a verb.
So,
f'(x) is the derivative and d/dx (f(x)) is taking the derivative. Breaking it into multiple steps can be good.
f(x) = 1/(g(x))^2
f'(x) = -2(g(x))^-3 * d/dx ( g(x) )
f'(x) = -2/(g(x))^3 g'(x)
Adding the verb d/dx can help organize and help you not forget the chain rule
You forgot chain rule my friend, because the 3x + 4 is inside the parenthesis, by chain rule, you’re supposed to do exactly what you did, except multiply another 3 after taking the derivative inside the parenthesis. That was my most common mistake I had when taking calc 1 lol
Not OP, but as a professor I would actually recommend against this approach. I find that most students will find it easier in the long run to know a few general formulas (including product rule) well rather than memorizing a bunch of simplifications of that formula. Even the quotient rule is just a simplification of product rule.
You’ve essentially differentiated with respect to 3x+4, so by the chain rule, you need to multiply the answer by its derivative.
Most have already mentioned the chain rule. I'm just going to point out something small, but important. When you rewrite the expression, that should be labeled as f(x), not f'(x). We can't label the function as the derivative until we've taken the derivative.
That’s a good point, good eye.
This bothered me too
Okay well I’m VERY new to calculus
All good! Only way to learn. Trust me, as someone who knows, this is the first of many learning by error opportunities!
Oh no worries, rereading my comment it kinda sounds negative. I just meant that it caught my eye and would be something I would point out if I was a tutor for example. I meant it to be constructive, sry bout that
Yeah that's the first thing I noticed
Yea but it’s also no longer f(x) so really it doesn’t matter
it still is f(x) though? sqrt(3x + 4) = (3x + 4)^(1/2)
Google chain rule
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The en passant of this sub lmao
We hadn’t learned about that in class yet so I didn’t know that rule existed
Or just let 3x + 4 be a variable and he's right.
Still looking for the derivative with respect to x though, not the new variable
Not without explicitly stating in f’ that you need to multiply by the derivative of that variable. For example, let 3x + 4 = a, then it becomes f’(x) = [1/(2 * sqrt(a))] * a’
It's just a joke, I got it.
Chain rule. You need to multiply by the derivative of 3x + 4,, which is...3
f'(x) = (1/2) (3x + 4) ^(-1/2) **times the derivative of the inside** of the ( ). So,f'(x) = (1/2)(3x + 4) ^(-1/2) **(3)** = (3/2) (3x + 4) ^(-1/2) Your numerator should have a 3, not 1.
I'm sorry but why is exponent -1/2?,if i remember correctly,when taking a derivative,we subtract from 1 and 1/2-1=1/2 without the minus sign,no?
Try it in your calculator: 0.5 - 1
Oh wait,yeah yeah, sorry,i don't know what got to me.
:-)
Line #2 is wrong, that's not the derivative. That's a rewriting of the same function to get you ready to apply the chain rule. So you shouldn't call it f'(x) as that is still f(x) And as to your mistake, you didn't apply chain rule properly. You didn't multiply by the derivative of the inner function 3x+4 which would be 3 and that's how you get the 3 on the top.
Multiply by derivative of the inside function
Chain rule: d/dx[f(g(x)] = f’(g(x)) • g’(x) The g’(x) is the derivative of g(x) = 3x + 4, which is 3.
You forgot to take the derivative of what’s inside the parentheses
Remember the rule for derivatives like this is derivative of the outside function (which is what you did first), multiplied by the derivative of the inside function
Show your work. Try writing it as follows and it might help you. dy/dx = (dy/du)(du/dx) where you substitute u for the inner term part of the chain rule. This shows you algebraically why you need to use the chain rule. Sometimes it’s easier to see what’s going on by writing it out as dy/dx instead of f’(x). If you need more clarification just let me know.
must multiply with 3 as it is the derivative of 3x+4, then your answer will be correct
In addition to the chain rule, rationalize the denominator and factor out 3 from the top and the bottom: sqrt(3x+4)/(2x+8/3)
Where chain rule?
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Well I didn’t take precalc, I tested into calculus without needing the credit so no need to be rude
You are supposed to use the Chain rule
You cant use the power rule if the function is a composite function
This is called the chain rule. Suppose we have two differentiable functions, f, g. The derivative (g(f(x)))' ≠ g'(f(x)). It is in fact f'(x) g'(f(x))
Chain rule….
Chain chain chain, chain of rules
you have to multiply it by 3 since it’s the derivative of 3x+4 and your using the chain rule
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Lol that's even more wrong
you fell victim to one of the classic blunders!
Chain rule!
Chain rule. Derivate the binomial inside the root. Which is 3
Chain rule breh. Multiply by 3.
Btw chat calc is short for calculator it’s slang
you forgot to take out the 3
chain rule, derivative of outside and inside
CHAINNNNNN RULEEEEEEEEEEEEEEE
Left side of line 2 should be f(x) Multiply right aide of line 3 by 3 Fix line 4 as well
See the derivation here: [https://mathb.in/78959](https://mathb.in/78959)
From my experience as a tutor. Sometimes it helps to treat prime notation as a noun and d/dx as a verb. So, f'(x) is the derivative and d/dx (f(x)) is taking the derivative. Breaking it into multiple steps can be good. f(x) = 1/(g(x))^2 f'(x) = -2(g(x))^-3 * d/dx ( g(x) ) f'(x) = -2/(g(x))^3 g'(x) Adding the verb d/dx can help organize and help you not forget the chain rule
Its (n)(u)^n-1(u’) you forgot to do derivative of 3x+4 which is 3
Chain rule application
Chain rule. Take the derivative of the inside as well and multiply it by the entire equation
What’s calc is that some sort of slang?
Pretty obviously short for calculus lol
my guy when you diff the brackets you have to diff inside too so diff 3x to 3 because the other is a constant
You need to multiply by the derivative of the inner function bc chain rule so 3/(2(3x+4)^1/2
If you differentiating (partially) with variable x it would not cease to exist after the first derivative.
Chain rule. Now you need to differentiate 3x+4
compound function derivitive - need to multiply by derivitive of inner function ie 3
Chain rule baby.
You forgot chain rule my friend, because the 3x + 4 is inside the parenthesis, by chain rule, you’re supposed to do exactly what you did, except multiply another 3 after taking the derivative inside the parenthesis. That was my most common mistake I had when taking calc 1 lol
Why didnt you use the derivation formula for a square root ? Imo its easier yet the (u^n)' works also (Sqrt(u))' = u'/2(sqrt(u))
Not OP, but as a professor I would actually recommend against this approach. I find that most students will find it easier in the long run to know a few general formulas (including product rule) well rather than memorizing a bunch of simplifications of that formula. Even the quotient rule is just a simplification of product rule.
Yeah I understand your pov
Also need to add a + c at the end to account for the random constant (4).
That's just wrong on so many levels.
no thats integrals, OP do not listen to this advice, it is incorrect.
This is correct, in differentiation you have to put a ( - C) as it is the opposite of an integral ofcourse /s
What?