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feeling dumb- differentiation

please see attached

It's been a while since I've looked at this. Is what I've written correct? (trying to make myself understand something). When I say 'other methods' I mean such as chain rule etc.
differentiation.png28.4KB
Reply 1
Avatar for krisshP
krisshP
14
First part right, but second part wrong.
6x + 2 is correct.

For the second part, you've got the part on the right (next to "to give") correct. You just have to multiply it by the derivative of the function inside the brackets.
The last one is not a function of x at all. It has a y in the bracket.
Reply 4
Avatar for ChocInABox
ChocInABox
OP
2
Original post by donutellme
6x + 2 is correct.

For the second part, you've got the part on the right (next to "to give") correct. You just have to multiply it by the derivative of the function inside the brackets.



Right so for the (x+2) ^2 and (4x+3)^4 I can't do what I did in the first one i.e. bringing the power to the front then reducing it by 1. This is because if we did this we would be differentiating with respect to x+2 and 4x+3 rather than x ? Consequently one must use chain rule for such questions?
(edited 9 years ago)
Original post by ChocInABox
Right so for the (x+2) ^2 and (4x+3)^4 I can't do what I did in the first one i.e. bringing the power to the front then reducing it by 1. This is because if we did this we would be differentiating with respect to x+2 and 4x+3 rather than x ? Consequently one must use chain rule for such questions?


You would use the chain rule for these questions, because these are a function of a function. So, just in case you don't know the chain rule:

- Treat the brackets like an x. Differentiate it and then put back the actual brackets.
- Multiply this by whatever the derivative of the bracket is.

e.g. for 4(2x2−3)34(2x^2-3)^3

- 4∗3(2x2−3)3−1∗(2∗2x2−1)4*3(2x^2-3)^{3-1} * (2*2x^{2-1})
- 12(2x2−3)2∗4x12(2x^2-3)^2 * 4x
- 48x(2x2−3)248x(2x^2-3)^2
Original post by ChocInABox
Right so for the (x+2) ^2 and (4x+3)^4 I can't do what I did in the first one i.e. bringing the power to the front then reducing it by 1. This is because if we did this we would be differentiating with respect to x+2 and 4x+3 rather than x ? Consequently one must use chain rule for such questions?


If you have a function of a function, you have to use the chain rule.

For instance the question you have is this:

y=(x2+1)12y = (x^2+1)^\frac{1}{2}

Let u=x2+1→dudx=2xu = x^2 + 1 \rightarrow \dfrac{du}{dx} = 2x.

Now, y=u12→dydu=12u−y = u^\frac{1}{2} \rightarrow \dfrac{dy}{du} = \frac{1}{2}u^-12=12(x2+1)−^\frac{1}{2} = \frac{1}{2}(x^2+1)^-12^\frac{1}{2}.

dydx=dydu×dudx\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}

That applies to all questions.
(edited 9 years ago)
Reply 7
Avatar for ChocInABox
ChocInABox
OP
2
oooooooooo right haa omg thank you that has cleared things up a lot I forgot about the whole 'function of a function' thing and so was getting confused when to use it. thank you both ! :biggrin:




Original post by donutellme
You would use the chain rule for these questions, because these are a function of a function. So, just in case you don't know the chain rule:

- Treat the brackets like an x. Differentiate it and then put back the actual brackets.
- Multiply this by whatever the derivative of the bracket is.

e.g. for 4(2x2−3)34(2x^2-3)^3

- 4∗3(2x2−3)3−1∗(2∗2x2−1)4*3(2x^2-3)^{3-1} * (2*2x^{2-1})
- 12(2x2−3)2∗4x12(2x^2-3)^2 * 4x
- 48x(2x2−3)248x(2x^2-3)^2


Original post by CTArsenal
If you have a function of a function, you have to use the chain rule.

For instance the question you have is this:

y=(x2+1)12y = (x^2+1)^\frac{1}{2}

Let u=x2+1→dudx=2xu = x^2 + 1 \rightarrow \dfrac{du}{dx} = 2x.

Now, y=u12→dydu=12u−y = u^\frac{1}{2} \rightarrow \dfrac{dy}{du} = \frac{1}{2}u^-12=12(x2+1)−^\frac{1}{2} = \frac{1}{2}(x^2+1)^-12^\frac{1}{2}.

dydx=dydu×dudx\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}

That applies to all questions.

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