What is the tangent of f(x)=(sqrt(2))^x,\at x=6 ?

The tangent of f(x)=(sqrt(2))^x,\at x=6 is y=4ln(2)x+8-24ln(2)

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tangent of f(x)=(sqrt(2))^x,\at x=6

tangent of

f (x) = (2)^{x}, at x = 6

y = 4 ln (2) x + 8 - 24 ln (2)

Solution steps

Find the tangent point:

(6, 8)

Find the slope of

f (x) = (2)^{x} : \frac{df ( x )}{d x} = \frac{1}{2} ln (2) e^{\frac{l n ( 2 ) x}{2}}

Compute the slope of

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

x = 6 : m = 4 ln (2)

Find the line with slope m=

4 ln (2)

and passing through

(6, 8) : y = 4 ln (2) x + 8 - 24 ln (2)

y = 4 ln (2) x + 8 - 24 ln (2)

y = e^{6 x} cos (π x), (0, 1)

y^{^{''}} + 14 y^{^{'}} + 28 y = 6675 e^{2 t} cos (15 t)

\frac{d}{d x} (5 x - 3)

t^{2} sin (t)

\frac{\partial}{\partial x} (x^{2} + y^{2} + 4 z^{2})

What is the tangent of f(x)=(sqrt(2))^x,\at x=6 ?
The tangent of f(x)=(sqrt(2))^x,\at x=6 is y=4ln(2)x+8-24ln(2)