If \( y = \frac{e^{-x} + e^x}{e^{-x} - e^x} \), then \( \frac{dy}{dx} \) is equal to:
\( \frac{4}{(e^{-x} - e^x)^2} \)
Finding the Derivative of an Exponential Function
The problem asks us to find the derivative of the function \( y = \frac{e^{-x} + e^x}{e^{-x} - e^x} \) with respect to \( x \). This type of function, which is a ratio of two functions involving exponential terms, is best differentiated using the quotient rule.
Understanding the Quotient Rule for Differentiation
The quotient rule is a fundamental rule in calculus for differentiating a function that is the ratio of two other differentiable functions. If we have a function \( y \) defined as \( y = \frac{u(x)}{v(x)} \), where \( u(x) \) and \( v(x) \) are differentiable functions of \( x \), then the derivative of \( y \) with respect to \( x \) is given by the formula:
\( \frac{dy}{dx} = \frac{v(x) \cdot \frac{du}{dx} - u(x) \cdot \frac{dv}{dx}}{(v(x))^2} \)
In simpler notation, using prime for derivatives (\( u' = \frac{du}{dx}, v' = \frac{dv}{dx} \)):
\( \frac{dy}{dx} = \frac{v u' - u v'}{v^2} \)
Applying the Quotient Rule to the Given Function
Our function is \( y = \frac{e^{-x} + e^x}{e^{-x} - e^x} \). Let's identify \( u(x) \) and \( v(x) \):
- \( u(x) = e^{-x} + e^x \) (the numerator)
- \( v(x) = e^{-x} - e^x \) (the denominator)
Now, we need to find the derivatives of \( u(x) \) and \( v(x) \) with respect to \( x \):
- Derivative of \( u(x) \):
\( \frac{du}{dx} = \frac{d}{dx}(e^{-x} + e^x) \) - Using the chain rule for \( e^{-x} \) (\( \frac{d}{dx} e^{f(x)} = e^{f(x)} f'(x) \)) and the standard derivative for \( e^x \) (\( \frac{d}{dx} e^x = e^x \)):
\( \frac{d}{dx}(e^{-x}) = e^{-x} \cdot \frac{d}{dx}(-x) = e^{-x} \cdot (-1) = -e^{-x} \)
\( \frac{d}{dx}(e^x) = e^x \)
So, \( u'(x) = -e^{-x} + e^x \) - Derivative of \( v(x) \):
\( \frac{dv}{dx} = \frac{d}{dx}(e^{-x} - e^x) \) - Using the derivatives found above:
\( \frac{dv}{dx} = -e^{-x} - e^x \) - So, \( v'(x) = -e^{-x} - e^x \)
Now, substitute \( u, v, u', \) and \( v' \) into the quotient rule formula:
\( \frac{dy}{dx} = \frac{(e^{-x} - e^x)(-e^{-x} + e^x) - (e^{-x} + e^x)(-e^{-x} - e^x)}{(e^{-x} - e^x)^2} \)
Simplifying the Expression for \( \frac{dy}{dx} \)
Let's simplify the numerator step-by-step:
Numerator \( = (e^{-x} - e^x)(-e^{-x} + e^x) - (e^{-x} + e^x)(-e^{-x} - e^x) \)
Simplify the first term in the numerator: \( (e^{-x} - e^x)(-e^{-x} + e^x) \)
- \( (e^{-x})(-e^{-x}) = -e^{-2x} \)
- \( (e^{-x})(e^x) = e^0 = 1 \)
- \( (-e^x)(-e^{-x}) = e^0 = 1 \)
- \( (-e^x)(e^x) = -e^{2x} \)
- So, the first term is \( -e^{-2x} + 1 + 1 - e^{2x} = 2 - e^{-2x} - e^{2x} \)
Simplify the second term in the numerator: \( (e^{-x} + e^x)(-e^{-x} - e^x) \)
Notice that \( (-e^{-x} - e^x) = -(e^{-x} + e^x) \). So the second term is \( (e^{-x} + e^x) \cdot -(e^{-x} + e^x) = -(e^{-x} + e^x)^2 \).
- Expanding \( -(e^{-x} + e^x)^2 \):
\( -((e^{-x})^2 + 2e^{-x}e^x + (e^x)^2) \) - \( -(e^{-2x} + 2e^0 + e^{2x}) \)
- \( -(e^{-2x} + 2 + e^{2x}) \)
- So, the second term is \( -e^{-2x} - 2 - e^{2x} \)
Now subtract the second term from the first term for the full numerator:
Numerator \( = (2 - e^{-2x} - e^{2x}) - (-e^{-2x} - 2 - e^{2x}) \)
Numerator \( = 2 - e^{-2x} - e^{2x} + e^{-2x} + 2 + e^{2x} \)
Notice that \( -e^{-2x} + e^{-2x} = 0 \) and \( -e^{2x} + e^{2x} = 0 \).
Numerator \( = 2 + 2 = 4 \)
The denominator is \( (e^{-x} - e^x)^2 \).
So, the derivative is:
\( \frac{dy}{dx} = \frac{4}{(e^{-x} - e^x)^2} \)
Matching the Result with the Options
Let's compare our calculated derivative with the given options:
- Option 1: \( \frac{4}{(e^{-x} - e^x)^2} \)
- Option 2: \( \frac{2}{(e^{-x} - e^x)^2} \)
- Option 3: \( \frac{-4}{(e^{-x} - e^x)^2} \)
- Option 4: \( \frac{-2}{(e^{-x} - e^x)^2} \)
Our result \( \frac{4}{(e^{-x} - e^x)^2} \) matches Option 1.
Revision Table: Differentiation Rules
Here's a quick summary of some basic differentiation rules used in calculus:
| Rule Name | Formula | Example |
|---|---|---|
| Constant Rule | \( \frac{d}{dx}(c) = 0 \) | \( \frac{d}{dx}(5) = 0 \) |
| Power Rule | \( \frac{d}{dx}(x^n) = nx^{n-1} \) | \( \frac{d}{dx}(x^3) = 3x^2 \) |
| Constant Multiple Rule | \( \frac{d}{dx}(cf(x)) = c\frac{d}{dx}(f(x)) \) | \( \frac{d}{dx}(2x^4) = 2(4x^3) = 8x^3 \) |
| Sum/Difference Rule | \( \frac{d}{dx}(f(x) \pm g(x)) = \frac{d}{dx}(f(x)) \pm \frac{d}{dx}(g(x)) \) | \( \frac{d}{dx}(x^2 + 3x) = 2x + 3 \) |
| Product Rule | \( \frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x) \) | \( \frac{d}{dx}(x e^x) = 1 \cdot e^x + x \cdot e^x = e^x(1+x) \) |
| Quotient Rule | \( \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2} \) | \( \frac{d}{dx}\left(\frac{\sin x}{x}\right) = \frac{x \cos x - \sin x \cdot 1}{x^2} \) |
| Chain Rule | \( \frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x) \) | \( \frac{d}{dx}(\sin(x^2)) = \cos(x^2) \cdot 2x \) |
| Exponential Rule (base e) | \( \frac{d}{dx}(e^x) = e^x \) | \( \frac{d}{dx}(e^{5x}) = e^{5x} \cdot 5 \) (using Chain Rule) |
Additional Information: Connecting to Hyperbolic Functions
The given function \( y = \frac{e^{-x} + e^x}{e^{-x} - e^x} \) can be related to hyperbolic functions.
Recall the definitions of hyperbolic cosine (\( \cosh x \)) and hyperbolic sine (\( \sinh x \)):
- \( \cosh x = \frac{e^x + e^{-x}}{2} \)
- \( \sinh x = \frac{e^x - e^{-x}}{2} \)
The hyperbolic cotangent (\( \coth x \)) is defined as \( \coth x = \frac{\cosh x}{\sinh x} \):
\( \coth x = \frac{\frac{e^x + e^{-x}}{2}}{\frac{e^x - e^{-x}}{2}} = \frac{e^x + e^{-x}}{e^x - e^{-x}} \)
Comparing this to our function \( y \):
\( y = \frac{e^{-x} + e^x}{e^{-x} - e^x} = \frac{e^x + e^{-x}}{-(e^x - e^{-x})} = -\frac{e^x + e^{-x}}{e^x - e^{-x}} = -\coth x \)
So, \( y = -\coth x \). Now let's find the derivative of \( y \) using the known derivative of \( \coth x \).
The derivative of \( \coth x \) is \( -\text{csch}^2 x \), where \( \text{csch} x = \frac{1}{\sinh x} \).
\( \frac{dy}{dx} = \frac{d}{dx}(-\coth x) = - \frac{d}{dx}(\coth x) = - (-\text{csch}^2 x) = \text{csch}^2 x \)
Now let's express \( \text{csch}^2 x \) in terms of exponentials:
\( \text{csch}^2 x = \left(\frac{1}{\sinh x}\right)^2 = \left(\frac{1}{\frac{e^x - e^{-x}}{2}}\right)^2 = \left(\frac{2}{e^x - e^{-x}}\right)^2 \)
\( \text{csch}^2 x = \frac{2^2}{(e^x - e^{-x})^2} = \frac{4}{(e^x - e^{-x})^2} \)
Since \( (e^x - e^{-x})^2 = (-(e^{-x} - e^x))^2 = (e^{-x} - e^x)^2 \), this result is identical to what we obtained using the quotient rule directly.
\( \frac{dy}{dx} = \frac{4}{(e^{-x} - e^x)^2} \)
This confirms our answer and shows the connection between exponential functions and hyperbolic functions in calculus.
Similar Questions
Differentiation of \( \log_5 (\log x^2) \) w.r.t. \( x \) is
The sum of values of \( a \) and \( b \) such that the function \( f(x) \) defined by
\[ f(x) = \begin{cases} 3, & x \leq 1 \\ ax + b, & 1 < x < 5 \\ 10, & x \geq 5 \end{cases} \] is a continuous function is
If \( f(x) = \begin{cases} \frac{\tan (\frac{\pi}{4} - x)}{\cot 2x}, & x \neq \frac{\pi}{4} \\ k, & x = \frac{\pi}{4} \end{cases} \) is continuous at \( x = \frac{\pi}{4} \), then the value of \( k \) will be equal to:
Important Questions from Continuity and Differentiability
Match List-I with List-II :
List-I
FunctionList-II
Derivative w.r.t. x(A) \( \frac{5^x}{\log_e 5} \) (I) \( 5^x (\log_e 5)^2 \) (B) \( \log_e 5 \) (II) \( 5^x \log_e 5 \) (C) \( 5^x \log_e 5 \) (III) \( 5^x \) (D) \( 5^x \) (IV) 0 Choose the correct answer from the options given below :
If f(x) is defined as:
\[ f(x) = \begin{cases} kx + 1 & \text{if } x \le \pi \\ \cos x & \text{if } x > \pi \end{cases} \]
is continuous at x = π, then the value of k is:
Differentiation of \( \log_5 (\log x^2) \) w.r.t. \( x \) is
The sum of values of \( a \) and \( b \) such that the function \( f(x) \) defined by
\[ f(x) = \begin{cases} 3, & x \leq 1 \\ ax + b, & 1 < x < 5 \\ 10, & x \geq 5 \end{cases} \] is a continuous function is
If \( f(x) = \begin{cases} \frac{\tan (\frac{\pi}{4} - x)}{\cot 2x}, & x \neq \frac{\pi}{4} \\ k, & x = \frac{\pi}{4} \end{cases} \) is continuous at \( x = \frac{\pi}{4} \), then the value of \( k \) will be equal to:
