What is the tangent of f(x)=x^2,\at x=1.9 ?

The tangent of f(x)=x^2,\at x=1.9 is y=3.8x-3.61

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tangent of f(x)=x^2,\at x=1.9

tangent of

f (x) = x^{2}, at x = 1.9

y = 3.8 x - 3.61

Solution steps

Find the tangent point:

(1.9, 3.61)

Find the slope of

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

Compute the slope of

f (x) = x^{2}

x = 1.9 : m = 3.8

Find the line with slope m=

3.8

and passing through

(1.9, 3.61) : y = 3.8 x - 3.61

y = 3.8 x - 3.61

\int \frac{1 - 0.4 x}{( 1 - x ) ( 5 - x )} d x

\int_{- \infty}^{\infty} t d t

\frac{d}{d x} ((5 - (4 - 3 x)^{2})^{3})

\frac{d y}{d x} = a^{x + y}

\int x^{4} tan^{3} (x^{5}) d x

What is the tangent of f(x)=x^2,\at x=1.9 ?
The tangent of f(x)=x^2,\at x=1.9 is y=3.8x-3.61