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3 The equation of the curve is
y=(e^(2x))/(3+4x).
(i) Find the coordinates of the stationary point on the curve, leaving your answer in exact value.

$3$ The equation of the curve is $y=\frac{e^{2 x}}{3+4 x}$ . $\newline$ (i) Find the coordinates of the stationary point on the curve, leaving your answer in exact value.

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Find derivatives using logarithmic differentiation

Full solution

Q.

3

The equation of the curve is

y=\frac{e^{2 x}}{3+4 x}

.

\newline

(i) Find the coordinates of the stationary point on the curve, leaving your answer in exact value.

Quotient Rule Application: To find the derivative, we use the quotient rule: $\frac{v(u') - u(v')}{v^2}$ , where $u = e^{2x}$ and $v = 3+4x$ .
Derivative of u: Differentiate $u$ with respect to $x$ : $u' = \frac{d}{dx}(e^{2x}) = 2e^{2x}$ .
Derivative of $v$ : Differentiate $v$ with respect to $x$ : $v' = \frac{d}{dx}(3+4x) = 4$ .
Numerator Simplification: Now apply the quotient rule: $y' = \frac{(3+4x)(2e^{2x}) - (e^{2x})(4)}{(3+4x)^2}$ .
Combine Like Terms: Simplify the numerator: $y' = \frac{6e^{2x} + 8xe^{2x} - 4e^{2x}}{(3+4x)^2}$ .
Stationary Points: Combine like terms in the numerator: $y' = \frac{2e^{2x} + 8xe^{2x}}{(3+4x)^2}$ .
Numerator Factorization: Set the derivative equal to zero to find the stationary points: $0 = \frac{(2e^{2x} + 8xe^{2x})}{(3+4x)^2}.$
Remaining Factor: The numerator must be zero for the fraction to be zero: $2e^{2x} + 8xe^{2x} = 0$ .
Solving for $x$ : Factor out $e^{2x}$ : $e^{2x}(2 + 8x) = 0$ .
Substitute $x$ into Equation: Since $e^{(2x)}$ is never zero, set the remaining factor equal to zero: $2 + 8x = 0$ .
Calculate Exponent: Solve for $x$ : $8^x = -2$ .
Calculate Denominator: Divide by $8$ : x = -rac{2}{8}.
Find y-coordinate: Simplify the fraction: $x = -\frac{1}{4}$ .
Stationary Point Coordinates: Now substitute $x = -\frac{1}{4}$ back into the original equation to find the $y$ -coordinate of the stationary point: $y = \frac{e^{2(-\frac{1}{4})}}{3+4(-\frac{1}{4})}$ .
Stationary Point Coordinates: Now substitute $x = -\frac{1}{4}$ back into the original equation to find the $y$ -coordinate of the stationary point: $y = \frac{e^{2(-\frac{1}{4})}}{3+4(-\frac{1}{4})}$ .Calculate the exponent: $e^{2(-\frac{1}{4})} = e^{-\frac{1}{2}}$ .
Stationary Point Coordinates: Now substitute $x = -\frac{1}{4}$ back into the original equation to find the $y$ -coordinate of the stationary point: $y = \frac{e^{2(-\frac{1}{4})}}{3+4(-\frac{1}{4})}$ .Calculate the exponent: $e^{2(-\frac{1}{4})} = e^{-\frac{1}{2}}$ .Calculate the denominator: $3+4(-\frac{1}{4}) = 3-1 = 2$ .
Stationary Point Coordinates: Now substitute $x = -\frac{1}{4}$ back into the original equation to find the y-coordinate of the stationary point: $y = \frac{e^{2(-\frac{1}{4})}}{3+4(-\frac{1}{4})}$ .Calculate the exponent: $e^{2(-\frac{1}{4})} = e^{-\frac{1}{2}}$ .Calculate the denominator: $3+4(-\frac{1}{4}) = 3-1 = 2$ .Now we have the y-coordinate: $y = \frac{e^{-\frac{1}{2}}}{2}$ .
Stationary Point Coordinates: Now substitute $x = -\frac{1}{4}$ back into the original equation to find the $y$ -coordinate of the stationary point: $y = \frac{e^{2(-\frac{1}{4})}}{3+4(-\frac{1}{4})}$ .Calculate the exponent: $e^{2(-\frac{1}{4})} = e^{-\frac{1}{2}}$ .Calculate the denominator: $3+4(-\frac{1}{4}) = 3-1 = 2$ .Now we have the $y$ -coordinate: $y = \frac{e^{-\frac{1}{2}}}{2}$ .The coordinates of the stationary point are $(-\frac{1}{4}, \frac{e^{-\frac{1}{2}}}{2})$ .

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