What is the tangent of x^2+x,\at x=1,x=4 ?

The tangent of x^2+x,\at x=1,x=4 is y=3x-1

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tangent of x^2+x,\at x=1,x=4

tangent of

x^{2} + x, at x = 1, x = 4

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} (\frac{24}{( x + 1 ) ^{5}})

\frac{1}{x}

y = 2 x, y = 9 x^{2}

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

\int sin (x) sin (2 x) sin (3 x) d x

What is the tangent of x^2+x,\at x=1,x=4 ?
The tangent of x^2+x,\at x=1,x=4 is y=3x-1