Question

What is the derivative at the point x=-2 on the curve $y=x^2$?
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Answer 1

We know that as per the first principle, the derivative of a function is given by $f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(X+h)-f\left(X\right)}{h}$
Let’s find it by first principle.
As per first principle
$f(x+h)={(x+h)}^2=x^2+h^2+2hxf\left(x\right)=x^2$
So $f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{x^2+h^2+2hx-x^2}{h}={lim}_{h\to 0}\frac{h^2+2hx}{h}={lim}_{h\to 0}\frac{h(h+2x)}{h}=2x$
So at x = -2, derivative will be 2 X (-2) = -4