What is the tangent of f(x)2x^2+8x,x=1 ?

The tangent of f(x)2x^2+8x,x=1 is y=(2(d/(d1)(f(1))+2f(1))+8)x-2f(1)-2 d/(d1)(f(1))

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tangent of f(x)2x^2+8x,x=1

tangent of

f (x) 2 x^{2} + 8 x, x = 1

y = (2 (\frac{d}{d 1} (f (1)) + 2 f (1)) + 8) x - 2 f (1) - 2 \frac{d}{d 1} (f (1))

Solution steps

Find the tangent point:

(1, 2 f (1) + 8)

Find the slope of

y = f (x) 2 x^{2} + 8 x : \frac{d y}{d x} = 2 (\frac{df}{d x} x^{2} + 2 x f (x)) + 8

Compute the slope of

y = f (x) 2 x^{2} + 8 x

x = 1 : m = 2 (\frac{d}{d 1} (f (1)) + 2 f (1)) + 8

Find the line with slope m=

2 (\frac{d}{d 1} (f (1)) + 2 f (1)) + 8

and passing through

(1, 2 f (1) + 8) : y = (2 (\frac{d}{d 1} (f (1)) + 2 f (1)) + 8) x - 2 f (1) - 2 \frac{d}{d 1} (f (1))

y = (2 (\frac{d}{d 1} (f (1)) + 2 f (1)) + 8) x - 2 f (1) - 2 \frac{d}{d 1} (f (1))

x \to 0 lim ((tan (6 x))^{x})

\frac{\partial}{\partial x} (x + y + x^{3} + y^{2})

y^{^{'}} = x^{2} (x - 2)^{2} (x - 1)^{2}

\int_{- 1}^{1} (3 x^{3} - x^{2} + x - 1) d x

x^{3} sin (7 x)

What is the tangent of f(x)2x^2+8x,x=1 ?
The tangent of f(x)2x^2+8x,x=1 is y=(2(d/(d1)(f(1))+2f(1))+8)x-2f(1)-2 d/(d1)(f(1))