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differentiation problem solving

8.PNG i need help on like how im supposed to format these into the lim h towards 0 formula. how i know i have to sub -1 for the x in tje (x+h) but how would i illustrate the 5x
so far i have done (-1+h)-5(-1)-(-1)^3-5(-1)/h is this right?
Original post by 13.oswin.tsang
so far i have done (-1+h)-5(-1)-(-1)^3-5(-1)/h is this right?


The gradient is

Unparseable latex formula:

$\begin{align*} \dfrac{f(-1+h)-f(-1)}{(-1+h)-(-1)} & = \dfrac{[(-1+h)^3 -5(-1+h)]-[(-1)^3 - 5(-1)]}{h} \\ & = \dfrac{(h-1)^3 - 5h +1}{h}\end{align*}$



So simplify the numerator now, and obtain the result.
That's very confusing but it's just x^3-5x but for the (x+h) part, you've just added the h
Original post by 13.oswin.tsang
That's very confusing but it's just x^3-5x but for the (x+h) part, you've just added the h


Why is it confusing??

Do you understand that the gradient is given by y1βˆ’y0x1βˆ’x0\dfrac{y_1 - y_0}{x_1 - x_0} ??

Here, {x1=βˆ’1+hx0=βˆ’1\begin{cases}x_1 = -1+h \\ x_0 = -1\end{cases}. Also {y1=f(βˆ’1+h)y0=f(βˆ’1)=4\begin{cases} y_1 = f(-1+h) \\ y_0 = f(-1) = 4 \end{cases}.

So, you do realise that f(βˆ’1+h)=(βˆ’1+h)3βˆ’5(βˆ’1+h)f(-1 + h) = (-1+h)^3 -5 (-1+h), right?

Hence the result follows: (βˆ’1+h)3βˆ’5h+1h\dfrac{(-1+h)^3 - 5h + 1}{h}
I understand it .it's just putting the numbers in the (x+h)-(x)h equation with a big number like x^3+10x
Original post by 13.oswin.tsang
I understand it .it's just putting the numbers in the (x+h)-(x)h equation with a big number like x^3+10x


Its not that bad, just expand the binomial.
ok thank you

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