What is the tangent of f(x)=x^2,\at x=1 ?

The tangent of f(x)=x^2,\at x=1 is y=2x-1

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tangent of f(x)=x^2,\at x=1

tangent of

f (x) = x^{2}, at x = 1

y = 2 x - 1

Solution steps

Find the tangent point:

(1, 1)

Find the slope of

f (x) = x^{2} : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2}

x = 1 : m = 2

Find the line with slope m=

2

and passing through

(1, 1) : y = 2 x - 1

y = 2 x - 1

y = 3 x

f (x) = ln (sinh (x))

sin (x)

y = (arcsin (x^{3}))^{4}

y = 3 x - 2

What is the tangent of f(x)=x^2,\at x=1 ?
The tangent of f(x)=x^2,\at x=1 is y=2x-1