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

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

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tangent of x^2,\at x=1

tangent of

x^{2}, at x = 1

y = 2 x - 1

Solution steps

Find the tangent point:

(1, 1)

Find the slope of

y = x^{2} : \frac{d y}{d x} = 2 x

Compute the slope of

y = 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

\frac{\partial}{\partial x} (\frac{1}{r ^{2}})

\int \frac{3}{e ^{5 x}} d x

f (x) = csc^{2} (x)

\int \frac{1}{1 - x} d x

\frac{d}{d x} (- 2 csc (x) + 4 cot (x))