Question

@asnaseer

Answer 1

do you know the shortcut rule?

Answer 2

what does your text book say about it?

Answer 3

one sec let me screenshot it

Answer 4

sorry just ignore this question, i dont see a shortcut method, i will have to ask my teacher

Answer 5

I can derive it for you if you want?

Answer 6

is that the shortcut method?

Answer 7

yes

Answer 8

i guess well do that then(:

Answer 9

:)

Answer 10

ok, you remember the limit definition gave us:\[f'(x)=\lim_{h\rightarrow 0}{\frac{f(x+h)-f(x)}{h}}\]

Answer 11

yes!

Answer 12

i got that down

Answer 13

now lets look at a special "family" fo functions defined as:\[f(x)=ax^b\]where "a" and "b" are just some constants.

Lets use the limit definition to see what we get...

Answer 14

\[f(x+h)=a(x+h)^b=a(x^b+bx^{b-1}h+\frac{b(b-1)}{2}x^{n-2}h^2+...+h^b)\]\[=ax^b+abx^{b-1}h+\frac{ab(b-1)}{2}x^{n-2}h^2+...+ah^b\]agreed?

Answer 15

*the \(x^{n-2}\) above should be \(x^{b-2}\)

Answer 16

I'm using the binomial expansion for \((x+h)^b\)

Answer 17

we then get:\[f(x+h)-f(x)=ax^b+abx^{b-1}h+\frac{ab(b-1)}{2}x^{b-2}h^2+...+ah^b-ax^b\]\[=abx^{b-1}h+\frac{ab(b-1)}{2}x^{b-2}h^2+...+ah^b\]\[=h(abx^{b-1}+\frac{ab(b-1)}{2}x^{b-2}h+...+ah^{b-1})\]

Answer 18

oh got it

Answer 19

this is the shortcut? hahaha seems a littl longer

Answer 20

so, finally we get:\[f'(x)=\lim_{h\rightarrow 0}{\frac{f(x+h)-f(x)}{h}}\]\[=\lim_{h\rightarrow 0}{\frac{h(abx^{b-1}+\frac{ab(b-1)}{2}x^{b-2}h+...+ah^{b-1})}{h}}\]
\[=\lim_{h\rightarrow 0}{abx^{b-1}+\frac{ab(b-1)}{2}x^{b-2}h+...+ah^{b-1}}\]

Answer 21

thank you so much!

Answer 22

it will become obvious in the next couple of steps :)

Answer 23

(:

Answer 24

now look at this final limit - can you see that all terms apart from the first one will become zero when h=0?

Answer 25

because every term except the first one contains an h

Answer 26

so the limit finally gives us:\[f'(x)=abx^{b-1}\]

Answer 27

we can therefore say that for all functions of the form:\[f(x)=ax^b\]we have:\[f'(x)=abx^{b-1}\]

Answer 28

you can similarly show that if a function is composed of the sum of expressions like this:\[f(x)=g_1(x)+g_2(x)+...+g_n(x)\]then:\[f'(x)=g_1'(x)+g_2'(x)+...+g_n'(x)\]

Answer 29

so given your function:\[f(x)=4.5x^2-3x+2=4.5x^2-3x^1+2\]we get, using these shortcut rules:\[f'(x)=4.5\times2\times x^{2-1}-3\times 1\times x^{1-0}\]\[=9x-3\]

Answer 30

I forgot to show you that there is shortcut rule that states that all constants become zero when differentiated. So the "2" in f(x) disappears when it is differentiated.

Answer 31

you can show that very simply. Lets say we have a function:\[f(x)=c\]where 'c' is just a constant, then we get:\[f'(x)=\lim_{h\rightarrow 0}{\frac{f(x+h)-f(x)}{h}}=\lim_{h\rightarrow 0}{\frac{c-c}{h}}=\lim_{h\rightarrow 0}{\frac{0}{h}}=0\]

Answer 32

I've shown you above how to apply these shortcut rules

Answer 33

The function you were given was:\[f(x)=4.5x^2-3x+2=4.5x^2-3x^1+2\]now look at each part, lets take \(4.5x^2\) first. Using the rules we know that if:\[f(x)=4.5x^2\]then:\[f'(x)=4.5\times2\times x^{2-1}=9x\]

Now lets look at the next term \(-3x\). Using the rules we know that if:\[f(x)=-3x=-3x^1\]then:\[f'(x)=-3\times1\times x^{1-1}=-3x^0=-3\]

Finally we are left with the const +2 term which we know just differentiates to zero

Answer 34

so, putting it all together we get:\[f(x)=4.5x^2-3x+2\]therefore:\[f'(x)=4.5\times2\times x^1-3\times1\times x^0+2=9x-3\]