Solve for dy/dx: y = (e^(mx) + e^(nx)) / (e^(mx) - e^(-nx))
Understand the Problem
The question provides an equation y = (e^(mx) + e^(nx)) / (e^(mx) - e^(-nx)) and asks us to solve it. Since the prompt is open ended, I will assume that they want us to find the derivative, dy/dx.
Answer
$$ \frac{dy}{dx} = \frac{-2n - (m+n)e^{(m-n)x}}{(e^{mx}-e^{-nx})^2}$$
Answer for screen readers
$$ \frac{dy}{dx} = \frac{-2n - (m+n)e^{(m-n)x}}{(e^{mx}-e^{-nx})^2}$$
Steps to Solve
- Rewrite the equation
The equation is:
$$y=\frac{e^{mx}+e^{nx}}{e^{mx}-e^{-nx}}$$
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Apply the quotient rule The quotient rule states that if $y = \frac{u}{v}$, then $\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$. Here, $u = e^{mx}+e^{nx}$ and $v = e^{mx}-e^{-nx}$.
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Find $\frac{du}{dx}$
$\frac{du}{dx} = \frac{d}{dx}(e^{mx}+e^{nx}) = me^{mx} + ne^{nx}$
- Find $\frac{dv}{dx}$
$\frac{dv}{dx} = \frac{d}{dx}(e^{mx}-e^{-nx}) = me^{mx} - (-n)e^{-nx} = me^{mx} + ne^{-nx}$
- Apply the quotient rule formula
$\frac{dy}{dx} = \frac{(e^{mx}-e^{-nx})(me^{mx} + ne^{nx}) - (e^{mx}+e^{nx})(me^{mx} + ne^{-nx})}{(e^{mx}-e^{-nx})^2}$
- Expand the numerator
Numerator: $$(e^{mx}-e^{-nx})(me^{mx} + ne^{nx}) - (e^{mx}+e^{nx})(me^{mx} + ne^{-nx})$$ $$= me^{2mx} + ne^{mx+nx} - me^{mx-nx} - ne^{0} - (me^{2mx} + ne^{mx-nx} + me^{mx+nx} + ne^{0})$$ $$= me^{2mx} + ne^{mx+nx} - me^{mx-nx} - n - me^{2mx} - ne^{mx-nx} - me^{mx+nx} - n$$ $$= - me^{mx-nx} - n - ne^{mx-nx} - n$$ $$= -2n - me^{mx-nx} - ne^{mx-nx} $$ $$= -2n - (m+n)e^{(m-n)x}$$
- Simplify the derivative
$$ \frac{dy}{dx} = \frac{-2n - (m+n)e^{(m-n)x}}{(e^{mx}-e^{-nx})^2}$$
$$ \frac{dy}{dx} = \frac{-2n - (m+n)e^{(m-n)x}}{(e^{mx}-e^{-nx})^2}$$
More Information
The derivative of the function was found using the quotient rule. Simplification involved expanding the terms in the numerator and combining like terms.
Tips
A common mistake is an error during the expansion and simplication of the numerator. Be careful to properly distribute the negative signs and to combine like terms at the end. Another common mistake is to forget the chain rule when taking the derivative of $e^{mx}$ or $e^{nx}$.
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