What is the tangent of f(x)=log_{6}(2x^2+4),\at x=4 ?

The tangent of f(x)=log_{6}(2x^2+4),\at x=4 is y= 4/(9ln(6))x+2-(16)/(9ln(6))

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tangent of f(x)=log_{6}(2x^2+4),\at x=4

tangent of

f (x) = lo g_{6} (2 x^{2} + 4), at x = 4

y = \frac{4}{9 ln ( 6 )} x + 2 - \frac{16}{9 ln ( 6 )}

Solution steps

Find the tangent point:

(4, 2)

Find the slope of

f (x) = lo g_{6} (2 x^{2} + 4) : \frac{df ( x )}{d x} = \frac{2 x}{ln ( 6 ) ( x ^{2} + 2 )}

Compute the slope of

f (x) = lo g_{6} (2 x^{2} + 4)

x = 4 : m = \frac{4}{9 ln ( 6 )}

Find the line with slope m=

\frac{4}{9 ln ( 6 )}

and passing through

(4, 2) : y = \frac{4}{9 ln ( 6 )} x + 2 - \frac{16}{9 ln ( 6 )}

y = \frac{4}{9 ln ( 6 )} x + 2 - \frac{16}{9 ln ( 6 )}

(- 1, - 5), (1, - 2)

e^{t} \cdot sin (2 t)

t \to 2 lim (sin (\frac{π}{4} t))

x \to 1 lim (\frac{x ^{2 n} - 2 x ^{n} + 1}{x ^{2 m} - 2 x ^{m} + 1})

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

What is the tangent of f(x)=log_{6}(2x^2+4),\at x=4 ?
The tangent of f(x)=log_{6}(2x^2+4),\at x=4 is y= 4/(9ln(6))x+2-(16)/(9ln(6))