What is the tangent of y=(sqrt(2))^x,\at x=8 ?

The tangent of y=(sqrt(2))^x,\at x=8 is y=8ln(2)x+16-64ln(2)

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tangent of y=(sqrt(2))^x,\at x=8

tangent of

y = (2)^{x}, at x = 8

y = 8 ln (2) x + 16 - 64 ln (2)

Solution steps

Find the tangent point:

(8, 16)

Find the slope of

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

Compute the slope of

y = (2)^{x}

x = 8 : m = 8 ln (2)

Find the line with slope m=

8 ln (2)

and passing through

(8, 16) : y = 8 ln (2) x + 16 - 64 ln (2)

y = 8 ln (2) x + 16 - 64 ln (2)

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

f^{^{'}} x = (2 x) (x^{4} - 8 x) + (x^{2}) (4 x^{3} - 8)

y = \frac{6 t - 3}{5 t + 8}

x \to \frac{π}{2} lim (sin (\frac{x}{2}))

n = 3 \sum \infty (\frac{e}{π})^{n}

What is the tangent of y=(sqrt(2))^x,\at x=8 ?
The tangent of y=(sqrt(2))^x,\at x=8 is y=8ln(2)x+16-64ln(2)