The value of derivative at the point $x = - 2$ on the curve $y = x^{2}$ is equal to .

-4

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

-2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A -4
We know that as per the first principle, the derivative of a function is given by $f^{'} (x) = lim h \to 0 \frac{f (x + h) - f (x)}{h}$

$f (x + h) = {(x + h)}^{2} = x^{2} + h^{2} + 2 h x f (x) = x^{2}$
So $f^{'} (x) = lim h \to 0 \frac{x^{2} + h^{2} + 2 h x - x^{2}}{h} = lim h \to 0 \frac{h^{2} + 2 h x}{h} = lim h \to 0 \frac{h (h + 2 x)}{h} = 2 x$
At x = -2, f'(x)= 2 $\times$ (-2) = -4

Suggest Corrections

Similar questions

What is the derivative at the point $x = - 2$ on the curve $y = x^{2}$ ?
___

Q. The tangents to the curve

y = x^{3} - 6 x^{2} + 9 x + 4, 0 \leq x \leq 5

has maximum slope at

x

which is equal to

If the tangent at A on the curve y= $x^{3}$ meets the curve again at B and the gradient at B is K times the gradient at A, then the value of K is

Q. The value of

s i n (c o t^{- 1} (c o s (t a n^{- 1} x)))

is equal to

If $A > 0, c, d, u, v$ are non-zero constants, and the graphs of $f (x) = | A x + c | + d$ and $g (x) = - | A x + u | + v$ intersect exactly at 2 points (1, 4) and (3, 1) then the value of $\frac{u + c}{A}$ equals to

Join BYJU'S Learning Program

The value of derivative at the point x=−2 on the curve y=x2 is equal to .

The correct option is A -4We know that as per the first principle, the derivative of a function is given by f′(x)=limh→0f(x+h)−f(x)h f(x+h)=(x+h)2=x2+h2+2hxf(x)=x2 So f′(x)=limh→0x2+h2+2hx−x2h=limh→0h2+2hxh=limh→0h(h+2x)h=2x At x = -2, f'(x)= 2 × (-2) = -4

The value of derivative at the point $x = - 2$ on the curve $y = x^{2}$ is equal to .