What is the tangent of f(x)=x^2+5x,\at x=4 ?

The tangent of f(x)=x^2+5x,\at x=4 is y=13x-16

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tangent of f(x)=x^2+5x,\at x=4

tangent of

f (x) = x^{2} + 5 x, at x = 4

y = 13 x - 16

Solution steps

Find the tangent point:

(4, 36)

Find the slope of

f (x) = x^{2} + 5 x : \frac{df ( x )}{d x} = 2 x + 5

Compute the slope of

f (x) = x^{2} + 5 x

x = 4 : m = 13

Find the line with slope m=

13

and passing through

(4, 36) : y = 13 x - 16

y = 13 x - 16

\int x e^{- x} - e^{- x} d x

\frac{\partial}{\partial x} (3 e^{3 x})

y^{^{'}} = y (1 - y), y (0) = 8

x cos (x) - sin (x)

\frac{1}{6} π r^{3}

What is the tangent of f(x)=x^2+5x,\at x=4 ?
The tangent of f(x)=x^2+5x,\at x=4 is y=13x-16