What is the tangent of f(x)= 8/(7x),\at 1,87 ?

The tangent of f(x)= 8/(7x),\at 1,87 is y=-8/(7a_{0)^2}x+(16)/(7a_{0)}

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tangent of f(x)= 8/(7x),\at 1,87

tangent of

f (x) = \frac{8}{7 x}, at 1, 87

y = - \frac{8}{7 a _{0}^{2}} x + \frac{16}{7 a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{8}{7 a _{0}})

Find the slope of

f (x) = \frac{8}{7 x} : \frac{df ( x )}{d x} = - \frac{8}{7 x ^{2}}

Compute the slope of

f (x) = \frac{8}{7 x}

x = a_{0} : m = - \frac{8}{7 a _{0}^{2}}

Find the line with slope m=

- \frac{8}{7 a _{0}^{2}}

and passing through

(a_{0}, \frac{8}{7 a _{0}}) : y = - \frac{8}{7 a _{0}^{2}} x + \frac{16}{7 a _{0}}

y = - \frac{8}{7 a _{0}^{2}} x + \frac{16}{7 a _{0}}

\frac{\partial}{\partial x} (x^{6} - 4 x^{2} y^{4} + 3 y^{2})

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

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

y^{^{''}} + 6 y^{^{'}} + 5 y = 16 t e^{5 t}

y^{^{''}} + 6 y^{^{'}} - 4 y = 0

What is the tangent of f(x)= 8/(7x),\at 1,87 ?
The tangent of f(x)= 8/(7x),\at 1,87 is y=-8/(7a_{0)^2}x+(16)/(7a_{0)}