What is the tangent of x^2+x,\at x=3 ?

The tangent of x^2+x,\at x=3 is y=7x-9

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tangent of x^2+x,\at x=3

tangent of

x^{2} + x, at x = 3

y = 7 x - 9

Solution steps

Find the tangent point:

(3, 12)

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 = 3 : m = 7

Find the line with slope m=

7

and passing through

(3, 12) : y = 7 x - 9

y = 7 x - 9

\int \frac{1}{x ^{2} + 6 x + 18} d x

\frac{d}{d x} (\frac{7}{2} x^{2} - 5 x + 1)

\frac{\partial}{\partial x} (cos^{2} (y))

y = \frac{x ^{2} - 1}{x ^{2} + x + 1}