What is the tangent of f(x)= x/((4x-3)^8),\at x=1 ?

The tangent of f(x)= x/((4x-3)^8),\at x=1 is y=-31x+32

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tangent of f(x)= x/((4x-3)^8),\at x=1

tangent of

f (x) = \frac{x}{( 4 x - 3 ) ^{8}}, at x = 1

y = - 31 x + 32

Solution steps

Find the tangent point:

(1, 1)

Find the slope of

f (x) = \frac{x}{( 4 x - 3 ) ^{8}} : \frac{df ( x )}{d x} = \frac{- 3 - 28 x}{( 4 x - 3 ) ^{9}}

Compute the slope of

f (x) = \frac{x}{( 4 x - 3 ) ^{8}}

x = 1 : m = - 31

Find the line with slope m=

- 31

and passing through

(1, 1) : y = - 31 x + 32

y = - 31 x + 32

8 - 2 x, x^{3} - 7 x^{2} + 12 x, 1, 4

\int \frac{x ^{4} + x + 1}{x ^{3} - x} d x

y^{^{'}} - y = 2 x

\frac{\partial}{\partial z} (2 x^{2} + 3 y^{2} - z^{2})

\frac{1}{( 2 x + 1 ) ^{2}}

What is the tangent of f(x)= x/((4x-3)^8),\at x=1 ?
The tangent of f(x)= x/((4x-3)^8),\at x=1 is y=-31x+32