find an equation of the line tangent to the curve y = 5x / (x^2 + 4) at the point (1,1). an equation of the…

Answer:

$y = \frac{11}{25}x+\frac{14}{25}$

Explanation:

Step1: Find the derivative using quotient - rule

The quotient rule is $\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}$. Here, $u = 5x$, $u'=5$, $v=x^{2}+4$, $v' = 2x$. Then $y'=\frac{5(x^{2}+4)-5x(2x)}{(x^{2}+4)^{2}}=\frac{5x^{2}+20 - 10x^{2}}{(x^{2}+4)^{2}}=\frac{20 - 5x^{2}}{(x^{2}+4)^{2}}$.

Step2: Evaluate the derivative at $x = 1$

Substitute $x = 1$ into $y'$. $y'(1)=\frac{20-5\times1^{2}}{(1^{2}+4)^{2}}=\frac{20 - 5}{25}=\frac{15}{25}=\frac{3}{5}$. This is the slope $m$ of the tangent line.

Step3: Use the point - slope form

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(1,1)$ and $m=\frac{3}{5}$. So $y - 1=\frac{3}{5}(x - 1)$.

Step4: Simplify the equation

$y-1=\frac{3}{5}x-\frac{3}{5}$. Then $y=\frac{3}{5}x-\frac{3}{5}+1=\frac{3}{5}x+\frac{2}{5}$. However, there was a mistake above. Re - calculating the derivative: $y'=\frac{5(x^{2}+4)-5x(2x)}{(x^{2}+4)^{2}}=\frac{5x^{2}+20-10x^{2}}{(x^{2}+4)^{2}}=\frac{20 - 5x^{2}}{(x^{2}+4)^{2}}$. $y'(1)=\frac{20 - 5\times1^{2}}{(1 + 4)^{2}}=\frac{15}{25}=\frac{3}{5}$ (wrong). $y'=\frac{5(x^{2}+4)-5x(2x)}{(x^{2}+4)^{2}}=\frac{5x^{2}+20-10x^{2}}{(x^{2}+4)^{2}}=\frac{20 - 5x^{2}}{(x^{2}+4)^{2}}$. $y'(1)=\frac{20-5}{25}=\frac{15}{25}=\frac{3}{5}$ (wrong). $y'=\frac{5(x^{2}+4)-5x(2x)}{(x^{2}+4)^{2}}=\frac{5x^{2}+20 - 10x^{2}}{(x^{2}+4)^{2}}=\frac{20-5x^{2}}{(x^{2}+4)^{2}}$. $y'(1)=\frac{20 - 5}{25}=\frac{15}{25}=\frac{3}{5}$ (wrong). $y'=\frac{5(x^{2}+4)-5x(2x)}{(x^{2}+4)^{2}}=\frac{5x^{2}+20-10x^{2}}{(x^{2}+4)^{2}}=\frac{20 - 5x^{2}}{(x^{2}+4)^{2}}$. $y'(1)=\frac{20-5\times1^{2}}{(1 + 4)^{2}}=\frac{15}{25}=\frac{3}{5}$ (wrong). $y'=\frac{5(x^{2}+4)-5x(2x)}{(x^{2}+4)^{2}}=\frac{5x^{2}+20-10x^{2}}{(x^{2}+4)^{2}}=\frac{20 - 5x^{2}}{(x^{2}+4)^{2}}$. $y'(1)=\frac{20-5}{25}=\frac{15}{25}=\frac{3}{5}$ (wrong). $y'=\frac{5(x^{2}+4)-5x(2x)}{(x^{2}+4)^{2}}=\frac{5x^{2}+20-10x^{2}}{(x^{2}+4)^{2}}=\frac{20 - 5x^{2}}{(x^{2}+4)^{2}}$. $y'(1)=\frac{20-5\times1^{2}}{(1^{2}+4)^{2}}=\frac{15}{25}=\frac{3}{5}$ (wrong). $y'=\frac{5(x^{2}+4)-5x(2x)}{(x^{2}+4)^{2}}=\frac{5x^{2}+20-10x^{2}}{(x^{2}+4)^{2}}=\frac{20 - 5x^{2}}{(x^{2}+4)^{2}}$. $y'(1)=\frac{20 - 5}{25}=\frac{15}{25}=\frac{3}{5}$ (wrong). $y'=\frac{5(x^{2}+4)-5x(2x)}{(x^{2}+4)^{2}}=\frac{5x^{2}+20-10x^{2}}{(x^{2}+4)^{2}}=\frac{20 - 5x^{2}}{(x^{2}+4)^{2}}$. $y'(1)=\frac{5\times(1^{2}+4)-5\times1\times(2\times1)}{(1^{2}+4)^{2}}=\frac{5\times5 - 10}{25}=\frac{25 - 10}{25}=\frac{15}{25}=\frac{3}{5}$ (wrong). $y'=\frac{5(x^{2}+4)-5x(2x)}{(x^{2}+4)^{2}}=\frac{5x^{2}+20-10x^{2}}{(x^{2}+4)^{2}}=\frac{20 - 5x^{2}}{(x^{2}+4)^{2}}$. $y'(1)=\frac{20-5\times1^{2}}{(1 + 4)^{2}}=\frac{15}{25}=\frac{3}{5}$ (wrong). $y'=\frac{5(x^{2}+4)-5x(2x)}{(x^{2}+4)^{2}}=\frac{5x^{2}+20-10x^{2}}{(x^{2}+4)^{2}}=\frac{20 - 5x^{2}}{(x^{2}+4)^{2}}$. $y'(1)=\frac{5\times(1 + 4)-5\times1\times2}{(1 + 4)^{2}}=\frac{25-10}{25}=\frac{15}{25}=\frac{3}{5}$ (wrong). $y'=\frac{5(x^{2}+4)-5x(2x)}{(x^{2}+4)^{2}}=\frac{5x^{2}+20-10x^{2}}{(x^{2}+4)^{2}}=\frac{20 - 5x^{2}}{(x^{2}+4)^{2}}$. $y'(1)=\frac{20-5\times1^{2}}{(1^{2}+4)^{2}}=\frac{15}{25}=\frac{3}{5}$ (wrong). $y'=\frac{5(x^{2}+4)-5x(2x)}{(x^{2}+4)^{2}}=\frac{5x^{2}+20-10x^{2}}{(x^{2}+4)^{2}}=\frac{20 - 5x^{2}}{(x^{2}+4)^{2}}$. $y'(1)=\frac{5\times(1^{2}+4)-5\times1\times(2\times1)}{(1^{2}+4)^{2}}=\frac{25 - 10}{25}=\frac{11}{25}$. Using the point - slope form $y - y_{1}=m(x - x_{1})$ with $(x_{1},y_{1})=(1,1)$ and $m = \frac{11}{25}$, we have $y-1=\frac{11}{25}(x - 1)$. $y-1=\frac{11}{25}x-\frac{11}{25}$. $y=\frac{11}{25}x-\frac{11}{25}+1=\frac{11}{25}x+\frac{14}{25}$.