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

The tangent of y=x^2+x,\at x=1 is y=3x-1

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tangent of y=x^2+x,\at x=1

tangent of

y = x^{2} + x, at x = 1

y = 3 x - 1

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

y = x^{2} + x : \frac{d y}{d x} = 2 x + 1

Compute the slope of

y = x^{2} + x

x = 1 : m = 3

Find the line with slope m=

3

and passing through

(1, 2) : y = 3 x - 1

y = 3 x - 1

\frac{d}{d x} (arctan (\frac{( a ^{2} - x ^{3} )}{b ^{2}}))

\frac{d}{d x} (\frac{1 - 2 x ^{2}}{x ^{4} - 2 x ^{2} + 6})

\int 1 + (4 x)^{2} d x

\frac{d}{d t} (2 t e^{- t})

\int \frac{1}{sin ( x ) cos ^{2} ( x )} d x

What is the tangent of y=x^2+x,\at x=1 ?
The tangent of y=x^2+x,\at x=1 is y=3x-1