What is the tangent of y=(sqrt(3))^x,\at x=6 ?

The tangent of y=(sqrt(3))^x,\at x=6 is y=(27ln(3))/2 x+27-81ln(3)

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tangent of y=(sqrt(3))^x,\at x=6

tangent of

y = (3)^{x}, at x = 6

y = \frac{27 ln ( 3 )}{2} x + 27 - 81 ln (3)

Solution steps

Find the tangent point:

(6, 27)

Find the slope of

y = (3)^{x} : \frac{d y}{d x} = \frac{1}{2} ln (3) e^{\frac{l n ( 3 ) x}{2}}

Compute the slope of

y = (3)^{x}

x = 6 : m = \frac{27 ln ( 3 )}{2}

Find the line with slope m=

\frac{27 ln ( 3 )}{2}

and passing through

(6, 27) : y = \frac{27 ln ( 3 )}{2} x + 27 - 81 ln (3)

y = \frac{27 ln ( 3 )}{2} x + 27 - 81 ln (3)

\int \frac{1}{( x + 5 ) ^{2} ( x - 4 )} d x

- 3 t

\frac{d}{d x} (6 sin (sin (x)))

\int sin^{3} (x co) s^{5} x d x

f (x) = tan (2 x) \cdot sin (2 x)

What is the tangent of y=(sqrt(3))^x,\at x=6 ?
The tangent of y=(sqrt(3))^x,\at x=6 is y=(27ln(3))/2 x+27-81ln(3)