What is the tangent of f(x)=4x^3sqrt(5x-1),\at x=1 ?

The tangent of f(x)=4x^3sqrt(5x-1),\at x=1 is y=29x-21

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tangent of f(x)=4x^3sqrt(5x-1),\at x=1

tangent of

f (x) = 4 x^{3} 5 x - 1, at x = 1

y = 29 x - 21

Solution steps

Find the tangent point:

(1, 8)

Find the slope of

f (x) = 4 x^{3} 5 x - 1 : \frac{df ( x )}{d x} = \frac{2 ( 35 x ^{3} - 6 x ^{2} )}{5 x - 1}

Compute the slope of

f (x) = 4 x^{3} 5 x - 1

x = 1 : m = 29

Find the line with slope m=

29

and passing through

(1, 8) : y = 29 x - 21

y = 29 x - 21

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

\frac{\partial}{\partial x} (e^{x} y^{2})

\frac{d x}{d t} = \frac{a}{x t}

\int - \frac{1}{2 x + 2} d x

x \to 2 - lim (\frac{6}{x - 2})

What is the tangent of f(x)=4x^3sqrt(5x-1),\at x=1 ?
The tangent of f(x)=4x^3sqrt(5x-1),\at x=1 is y=29x-21