What is the tangent of f(x)=3x^2-ln(x),(1,3) ?

The tangent of f(x)=3x^2-ln(x),(1,3) is y=5x-2

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tangent of f(x)=3x^2-ln(x),(1,3)

tangent of

f (x) = 3 x^{2} - ln (x), (1, 3)

y = 5 x - 2

Solution steps

Find the slope of

f (x) = 3 x^{2} - ln (x) : \frac{df ( x )}{d x} = 6 x - \frac{1}{x}

Compute the slope of

f (x) = 3 x^{2} - ln (x)

(1, 3) : m = 5

Find the line with slope m=

5

and passing through

(1, 3) : y = 5 x - 2

y = 5 x - 2

f (x) = x^{5}, at x = 4

\int (x^{2} + 1) cos (2 x) d x

\int sin (2 x) e^{c o s (2 x)} d x

\int 7 x^{2} + 3 + \frac{6}{x ^{2} + 1} d x

\frac{d}{d x} (\frac{16}{( 5 - x ) ^{2}} - 1)

What is the tangent of f(x)=3x^2-ln(x),(1,3) ?
The tangent of f(x)=3x^2-ln(x),(1,3) is y=5x-2