What is the tangent of (5x)/(x+4)(1.1) ?

The tangent of (5x)/(x+4)(1.1) is y=(22)/((a_{0)+4)^2}x+(5.5a_{0}^2)/((a_{0)+4)^2}

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tangent of (5x)/(x+4)(1.1)

tangent of

\frac{5 x}{x + 4} (1.1)

y = \frac{22}{( a _{0} + 4 ) ^{2}} x + \frac{5.5 a _{0}^{2}}{( a _{0} + 4 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{5.5 a _{0}}{a _{0} + 4})

Find the slope of

y = \frac{5 x}{x + 4} 1.1 : \frac{d y}{d x} = \frac{22}{( x + 4 ) ^{2}}

Compute the slope of

y = \frac{5 x}{x + 4} 1.1

x = a_{0} : m = \frac{22}{( a _{0} + 4 ) ^{2}}

Find the line with slope m=

\frac{22}{( a _{0} + 4 ) ^{2}}

and passing through

(a_{0}, \frac{5.5 a _{0}}{a _{0} + 4}) : y = \frac{22}{( a _{0} + 4 ) ^{2}} x + \frac{5.5 a _{0}^{2}}{( a _{0} + 4 ) ^{2}}

y = \frac{22}{( a _{0} + 4 ) ^{2}} x + \frac{5.5 a _{0}^{2}}{( a _{0} + 4 ) ^{2}}

y^{^{''}} + 5 y^{^{'}} = e^{- t}

y^{^{'}} + y cos (x) = 2 cos (x)

y = \frac{10}{x ^{2} - 2 x - 24}, x = - 2, x = 2

h^{^{'}} = 6 - 5 cos (\frac{π}{15} t)

\int \frac{1 - sin ( x )}{x + cos ( x )} d x

What is the tangent of (5x)/(x+4)(1.1) ?
The tangent of (5x)/(x+4)(1.1) is y=(22)/((a_{0)+4)^2}x+(5.5a_{0}^2)/((a_{0)+4)^2}