What is the tangent of f(x)=(6x)/((5-2x)^4),\at x=1 ?

The tangent of f(x)=(6x)/((5-2x)^4),\at x=1 is y= 22/81 x-16/81

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tangent of f(x)=(6x)/((5-2x)^4),\at x=1

tangent of

f (x) = \frac{6 x}{( 5 - 2 x ) ^{4}}, at x = 1

y = \frac{22}{81} x - \frac{16}{81}

Solution steps

Find the tangent point:

(1, \frac{2}{27})

Find the slope of

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

Compute the slope of

f (x) = \frac{6 x}{( 5 - 2 x ) ^{4}}

x = 1 : m = \frac{22}{81}

Find the line with slope m=

\frac{22}{81}

and passing through

(1, \frac{2}{27}) : y = \frac{22}{81} x - \frac{16}{81}

y = \frac{22}{81} x - \frac{16}{81}

x 1 + y^{2} d x = d y

\frac{d}{d x} (\frac{ln ( x )}{5 x ^{3}})

\int 7 x sec^{2} (x) d x

10000 - 9000 e^{- 0.3 t}

\frac{s}{( s - 1 ) ^{2}}

What is the tangent of f(x)=(6x)/((5-2x)^4),\at x=1 ?
The tangent of f(x)=(6x)/((5-2x)^4),\at x=1 is y= 22/81 x-16/81