Infinite Diferential

Note the identity $1 + x + x^2 + \ldots = \dfrac{1}{1-x},$ which follows from the formula for a geometric series. (Since $x = 1/2,$ the sum converges.) Then, we have, by the Quotient Rule, $\begin{aligned} \dfrac{d}{dx}\left(\dfrac{1}{1-x}\right) &= \dfrac{1(1) - (1-x)(0)}{(1-x)^2} \\ &= \dfrac{1}{(1-x)^2} \end{aligned}$ so when $x = 1/2,$ the gradient of $y$ is $\dfrac{1}{(1-1/2)^2} = \boxed{4}.$

To find the gradient, we have to differentiate the equation:

$y=1+x+x^2+x^3 \dots x^n \dots$

$\frac {dy}{dx} = 0+1+2x+3x^2 \dots nx^{n-1} \dots$

Then, we substitute $x = \frac {1}{2}$ and solve the resulting geometric series:

$\frac {dy}{dx} = 1+2(\frac {1}{2})+3(\frac {1}{2})^2 \dots n(\frac {1}{2})^{n-1} \dots$

$2 \times \frac {dy}{dx} = 2+2+3(\frac {1}{2})+4(\frac {1}{2})^2 \dots n(\frac {1}{2})^{n-2} \dots$

We subtract the first equation from the second equation:

$\frac {dy}{dx} = 3+\frac {1}{2}+(\frac {1}{2})^2+(\frac {1}{2})^3 \dots (\frac {1}{2})^n \dots =3+1 = \boxed 4$

by differentiating we get 1+2x+3x^2+................................+nx^(n-1) by binomial theorem it is equal to (1-x)^(-2) putting x =1/2 we get 4

y = "unlimited pattern" 1/1-x = 1/1-1/2 = 1/(1/2) = 2. Because gradient (m) = y/x. So, the gradient is 2/(1/2)=4. Answer : 4

$y=1+x+x^2+x^3+ \cdots + x^n + \cdots \infty=\frac{1}{1-x} \Rightarrow y' = \frac{1}{(1-x)^2} \Rightarrow y'(\frac{1}{2}) = \boxed{4}$ .

However, $y$ as shown in the question is an infinite power series. I am unsure if it can be treated as a function. At most we can say, $y'(\frac{1}{2}) \rightarrow 4 \text{ as } n \rightarrow \infty$ .

dy=1+2x+3x^2+4x^3+................+...........to infinity. After Summing up the series we will get 1/(1-x)^2.Put x=1/2 and get the ans.

The sequence can be write down as $y= 1/(1-x)$ , and the gradient can be found by finding the derivative of this function.

The derivative: $y'= 1/(1-x)^2$

Put $x=1/2$ to the derivative to get a gradient: $y'= 4$