find an equation of the tangent line to the curve $y = \\frac{9e^{x}}{(1 + x^{2})}$ at the point…

find an equation of the tangent line to the curve $y = \\frac{9e^{x}}{(1 + x^{2})}$ at the point $(1,\\frac{9}{2}e)$. solution according to the quotient rule, we have $\\frac{dy}{dx}=\\frac{(1 + x^{2})\\frac{d}{dx}(9e^{x})-9e^{x}\\frac{d}{dx}(1 + x^{2})}{(1 + x^{2})^{2}}=\\frac{(1 + x^{2})(9e^{x})-9e^{x}(2x)}{(1 + x^{2})^{2}}=\\frac{9e^{x}(x^{2}-2x + 1)}{(1 + x^{2})^{2}}$. so the slope of the tangent line at $(1,\\frac{9}{2}e)$ is $\\frac{dy}{dx}\\big|_{x = 1}=0$. this means that the tangent line at $(1,\\frac{9}{2}e)$ is horizontal and its equation is $y = \\frac{9}{2}e$.

find an equation of the tangent line to the curve $y = \\frac{9e^{x}}{(1 + x^{2})}$ at the point $(1,\\frac{9}{2}e)$. solution according to the quotient rule, we have $\\frac{dy}{dx}=\\frac{(1 + x^{2})\\frac{d}{dx}(9e^{x})-9e^{x}\\frac{d}{dx}(1 + x^{2})}{(1 + x^{2})^{2}}=\\frac{(1 + x^{2})(9e^{x})-9e^{x}(2x)}{(1 + x^{2})^{2}}=\\frac{9e^{x}(x^{2}-2x + 1)}{(1 + x^{2})^{2}}$. so the slope of the tangent line at $(1,\\frac{9}{2}e)$ is $\\frac{dy}{dx}\\big|_{x = 1}=0$. this means that the tangent line at $(1,\\frac{9}{2}e)$ is horizontal and its equation is $y = \\frac{9}{2}e$.

Answer

Explanation:

Step1: Apply quotient - rule

The quotient rule for $y=\frac{u}{v}$ is $\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^{2}}$. Here, $u = 9e^{x}$ and $v=1 + x^{2}$. So, $\frac{dy}{dx}=\frac{(1 + x^{2})\frac{d}{dx}(9e^{x})-9e^{x}\frac{d}{dx}(1 + x^{2})}{(1 + x^{2})^{2}}$.

Step2: Differentiate $u$ and $v$

We know that $\frac{d}{dx}(9e^{x})=9e^{x}$ and $\frac{d}{dx}(1 + x^{2}) = 2x$. Substituting these into the quotient - rule formula, we get $\frac{dy}{dx}=\frac{(1 + x^{2})\times9e^{x}-9e^{x}\times2x}{(1 + x^{2})^{2}}$.

Step3: Simplify the derivative

Factor out $9e^{x}$ from the numerator: $\frac{dy}{dx}=\frac{9e^{x}(x^{2}-2x + 1)}{(1 + x^{2})^{2}}$.

Step4: Find the slope at $x = 1$

Substitute $x = 1$ into $\frac{dy}{dx}$: $\left.\frac{dy}{dx}\right|_{x = 1}=\frac{9e^{1}(1^{2}-2\times1 + 1)}{(1+1^{2})^{2}}=\frac{9e(1 - 2 + 1)}{4}=0$.

Step5: Find the equation of the tangent line

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(1,\frac{9}{2}e)$ and $m = 0$. Substituting these values, we get $y-\frac{9}{2}e=0\times(x - 1)$, so $y=\frac{9}{2}e$.

Answer:

$y=\frac{9}{2}e$