What is the tangent of f(x)=log_{5}(2x^2+7),\at x=3 ?

The tangent of f(x)=log_{5}(2x^2+7),\at x=3 is y=(12)/(25ln(5))x+2-(36)/(25ln(5))

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tangent of f(x)=log_{5}(2x^2+7),\at x=3

tangent of

f (x) = lo g_{5} (2 x^{2} + 7), at x = 3

y = \frac{12}{25 ln ( 5 )} x + 2 - \frac{36}{25 ln ( 5 )}

Solution steps

Find the tangent point:

(3, 2)

Find the slope of

f (x) = lo g_{5} (2 x^{2} + 7) : \frac{df ( x )}{d x} = \frac{4 x}{ln ( 5 ) ( 2 x ^{2} + 7 )}

Compute the slope of

f (x) = lo g_{5} (2 x^{2} + 7)

x = 3 : m = \frac{12}{25 ln ( 5 )}

Find the line with slope m=

\frac{12}{25 ln ( 5 )}

and passing through

(3, 2) : y = \frac{12}{25 ln ( 5 )} x + 2 - \frac{36}{25 ln ( 5 )}

y = \frac{12}{25 ln ( 5 )} x + 2 - \frac{36}{25 ln ( 5 )}

\frac{d}{d x} (sin (x) + 2 sin^{2} (x))

1 0^{x}, 1

\frac{d}{d x} (- 8 (64 e^{- 4 x^{2}} x^{3} - 24 e^{- 4 x^{2}} x))

y = 11 x^{2} e^{x}

\int_{1}^{7} \frac{1}{x} d x

What is the tangent of f(x)=log_{5}(2x^2+7),\at x=3 ?
The tangent of f(x)=log_{5}(2x^2+7),\at x=3 is y=(12)/(25ln(5))x+2-(36)/(25ln(5))