What is the tangent of f(x)=4e^x-x^{16}x,\at x=0 ?

The tangent of f(x)=4e^x-x^{16}x,\at x=0 is y=4x+4

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tangent of f(x)=4e^x-x^{16}x,\at x=0

tangent of

f (x) = 4 e^{x} - x^{16} x, at x = 0

y = 4 x + 4

Solution steps

Find the tangent point:

(0, 4)

Find the slope of

f (x) = 4 e^{x} - x^{16} x : \frac{df ( x )}{d x} = 4 e^{x} - 17 x^{16}

Compute the slope of

f (x) = 4 e^{x} - x^{16} x

x = 0 : m = 4

Find the line with slope m=

4

and passing through

(0, 4) : y = 4 x + 4

y = 4 x + 4

\frac{\partial}{\partial x} (π \cdot (\frac{x}{2})^{2} h)

\frac{1}{4 x ^{2} + 3 x}

\int 36 e^{4 x} 3 + e^{4 x} d x

\frac{d}{d y} (arctan (y))

sin (x) y^{^{'}} = y + 2 x

What is the tangent of f(x)=4e^x-x^{16}x,\at x=0 ?
The tangent of f(x)=4e^x-x^{16}x,\at x=0 is y=4x+4