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Help with chain-rule question? y=[(x+3)^4]-4x

How do I differentiate y=[(x+3)^4]-4x
I know how to use the chain rule, but what do I do with the -4x bit? If t=x+3, should i differentiate y=(t^3)-4x?
Reply 1
ddxy=ddx((x+3)4)ddx(4x)\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}y = \frac{\mathrm{d}}{\mathrm{d}x} \bigg( (x+3)^4 \bigg) - \frac{\mathrm{d}}{\mathrm{d}x}(4x).

The differential operator is linear, the derivative of a sum of function is the sum of the derivatives of each individual function.
Original post by yoshimatey
How do I differentiate y=[(x+3)^4]-4x
I know how to use the chain rule, but what do I do with the -4x bit? If t=x+3, should i differentiate y=(t^3)-4x?

This video should explain everything:
http://www.examsolutions.net/maths-revision/core-maths/differentiation/methods/product-rule/tutorial-1.php
Reply 3
Original post by yoshimatey
How do I differentiate y=[(x+3)^4]-4x
I know how to use the chain rule, but what do I do with the -4x bit? If t=x+3, should i differentiate y=(t^3)-4x?


the -4x is a separate term
Reply 4
Original post by yoshimatey
How do I differentiate y=[(x+3)^4]-4x
I know how to use the chain rule, but what do I do with the -4x bit? If t=x+3, should i differentiate y=(t^3)-4x?


The chain rule, (d/Dx)((x+3)^4)=(du^4)/du) (du/dx) where u=x+3 and (d/du) u^4= 4u^3


Posted from TSR Mobile
(edited 8 years ago)
just differentiate each bit separately ?
Reply 6
Original post by Zacken
ddxy=ddx((x+3)4)ddx(4x)\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}y = \frac{\mathrm{d}}{\mathrm{d}x} \bigg( (x+3)^4 \bigg) - \frac{\mathrm{d}}{\mathrm{d}x}(4x).

The differential operator is linear, the derivative of a sum of function is the sum of the derivatives of each individual function.


THANKS A BILLION! You have really helped me out! :smile:
Reply 7
Original post by yoshimatey
THANKS A BILLION! You have really helped me out! :smile:


You're very welcome! :smile:

You can even check this work for certain functions y=3x=x+x+xy = 3x = x + x + x, so dydx=ddxx+ddxx+ddxx=1+1+1=3\frac{dy}{dx} = \frac{\mathrm{d}}{\mathrm{d}x}x + \frac{\mathrm{d}}{\mathrm{d}x}x + \frac{\mathrm{d}}{\mathrm{d}x}x = 1 + 1 + 1 = 3

But also, ddxy=ddx(3x)=3\frac{\mathrm{d}}{\mathrm{d}x}y = \frac{\mathrm{d}}{\mathrm{d}x}(3x) = 3
Original post by Zacken
You're very welcome! :smile:

You can even check this work for certain functions y=3x=x+x+xy = 3x = x + x + x, so dydx=ddxx+ddxx+ddxx=1+1+1=3\frac{dy}{dx} = \frac{\mathrm{d}}{\mathrm{d}x}x + \frac{\mathrm{d}}{\mathrm{d}x}x + \frac{\mathrm{d}}{\mathrm{d}x}x = 1 + 1 + 1 = 3

But also, ddxy=ddx(3x)=3\frac{\mathrm{d}}{\mathrm{d}x}y = \frac{\mathrm{d}}{\mathrm{d}x}(3x) = 3

Or prove it.
Reply 9
Original post by morgan8002
Or prove it.


Indeed, although it might be a little soon to ask the person to prove it given their unfamiliarity with limits, but for anybody interested:

Let h(x)=f(x)+g(x)h(x) = f(x) + g(x), then

h(x)=lima0h(x+a)h(x)a\displaystyle h'(x) = \lim_{a \to 0} \frac{h(x+a) - h(x)}{a}

=lima0f(x+a)+g(x+a)f(x)g(x)a\displaystyle = \lim_{a \to 0}\frac{f(x+a) + g(x+a) - f(x) - g(x)}{a}

and a bit of shuffling and the sum rule for limits yields


h(x)=lima0f(x+a)f(x)a+lima0g(x+a)g(x)a=f(x)+g(x)\displaystyle h'(x) = \lim_{a \to 0} \frac{f(x+a) - f(x)}{a} + \lim_{a \to 0}\frac{g(x+a) - g(x)}{a} = f'(x) + g'(x)
Reply 10
Original post by Zacken
Indeed, although it might be a little soon to ask the person to prove it given their unfamiliarity with limits, but for anybody interested:

Let h(x)=f(x)+g(x)h(x) = f(x) + g(x), then

h(x)=lima0h(x+a)h(x)a\displaystyle h'(x) = \lim_{a \to 0} \frac{h(x+a) - h(x)}{a}

=lima0f(x+a)+g(x+a)f(x)g(x)a\displaystyle = \lim_{a \to 0}\frac{f(x+a) + g(x+a) - f(x) - g(x)}{a}

and a bit of shuffling and the sum rule for limits yields


h(x)=lima0f(x+a)f(x)a+lima0g(x+a)g(x)a=f(x)+g(x)\displaystyle h'(x) = \lim_{a \to 0} \frac{f(x+a) - f(x)}{a} + \lim_{a \to 0}\frac{g(x+a) - g(x)}{a} = f'(x) + g'(x)


'overkill mate' - Ronald Weasley

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