Question

The slope of tangent on a curve y= 2x^2 - 1 at x=2 is

explanation needed with answer​

Answer 1

AnsWer :

8.

Solution :

we have,

$\tt \dagger \: \: y = 2 {x}^{2} - 1.$

Now,

$\tt\longmapsto\frac{dy}{dx} = \tan \theta$

Let's, Differentiate y With respect to x ( W.r.t.x.)

$\rule{80}1$

$\tt\longmapsto \frac{dy}{dx} = 2 {x}^{2} - 1.$

$\tt\longmapsto \frac{dy}{dx} = 2.2x - 0$

$\tt\longmapsto \frac{dy}{dx} = 4x.$

$\rule{30}1$

Now, Putting the value of x.

$\tt\longmapsto \frac{dy}{dx} = 4 \times 2.$

$\tt\longmapsto \frac{dy}{dx} = 8.$

$\rule{40}1$

Note : Graph provide in attachment.

Therefore, the value of slop of tangent on curve be 8.

Answer 2

Answer:

y = 2x²-1 (given)

slope is 8.

Explanation:

for slope

f(x)=2x²-1

f'(x)=4x

given,

x=2

f'(x) =4×2

=8.

then slope is 8

i hope it is helpful for you.