What is the derivative of e xwith respect to x e?
Understanding the Calculus Problem: Derivative with Respect to a Function
The question asks for the derivative of one function (\(e^x\)) with respect to another function (\(x^e\)). This is different from the standard derivative with respect to the variable \(x\). When we need to find the derivative of a function \(y\) with respect to another function \(u\), we can use the formula derived from the chain rule:
Let \(y = f(x)\) and \(u = g(x)\). We want to find \(\frac{dy}{du}\). We can express this as: \[ \frac{dy}{du} = \frac{\frac{dy}{dx}}{\frac{du}{dx}} \]
In this specific problem, we have:
- \(y = e^x\)
- \(u = x^e\)
We need to calculate the derivative of \(y\) with respect to \(x\) (\(\frac{dy}{dx}\)) and the derivative of \(u\) with respect to \(x\) (\(\frac{du}{dx}\)), and then divide the first by the second.
Step-by-Step Solution to Find the Derivative
Step 1: Find the derivative of \(y = e^x\) with respect to \(x\)
The derivative of the exponential function \(e^x\) with respect to \(x\) is a standard differentiation formula.
\[ \frac{dy}{dx} = \frac{d}{dx}(e^x) \] \[ \frac{dy}{dx} = e^x \]Step 2: Find the derivative of \(u = x^e\) with respect to \(x\)
The function \(u = x^e\) is a power function where the base is the variable \(x\) and the exponent is a constant (\(e\) is a mathematical constant, approximately 2.718). The power rule for differentiation states that the derivative of \(x^n\) with respect to \(x\) is \(nx^{n-1}\), where \(n\) is a constant.
Applying the power rule with \(n=e\):
\[ \frac{du}{dx} = \frac{d}{dx}(x^e) \] \[ \frac{du}{dx} = e \cdot x^{e-1} \]Step 3: Calculate the derivative of \(y\) with respect to \(u\)
Now we use the formula \(\frac{dy}{du} = \frac{\frac{dy}{dx}}{\frac{du}{dx}}\). Substitute the derivatives we found in Step 1 and Step 2:
\[ \frac{dy}{du} = \frac{e^x}{ex^{e-1}} \]Step 4: Simplify the expression
The expression we obtained is \(\frac{e^x}{ex^{e-1}}\). We can rewrite \(x^{e-1}\) using the property of exponents \(a^{m-n} = \frac{a^m}{a^n}\).
\[ x^{e-1} = \frac{x^e}{x^1} = \frac{x^e}{x} \]Now substitute this back into the expression for \(\frac{dy}{du}\):
\[ \frac{dy}{du} = \frac{e^x}{e \cdot \left(\frac{x^e}{x}\right)} \] \[ \frac{dy}{du} = \frac{e^x}{\frac{ex^e}{x}} \]To divide by a fraction, we multiply by its reciprocal:
\[ \frac{dy}{du} = e^x \cdot \frac{x}{ex^e} \] \[ \frac{dy}{du} = \frac{xe^x}{ex^e} \]This is the final simplified derivative of \(e^x\) with respect to \(x^e\).
Comparing the Result with Options
Let's compare our derived result, \(\dfrac{xe^x}{ex^e}\), with the given options:
| Option | Expression | Matches our result? |
|---|---|---|
| 1 | \(\dfrac{xe^x}{ex^e}\) | Yes |
| 2 | \(\dfrac{e^x}{x^e}\) | No |
| 3 | \(\dfrac{xe^x}{x^e}\) | No |
| 4 | \(\dfrac{e^x}{ex^e}\) | No |
Our calculated derivative matches Option 1.
Revision Table: Key Differentiation Formulas
Here are some fundamental differentiation formulas used in calculus:
| Function \(f(x)\) | Derivative \(f'(x) = \frac{d}{dx}(f(x))\) |
|---|---|
| \(c\) (constant) | \(0\) |
| \(x^n\) | \(nx^{n-1}\) |
| \(e^x\) | \(e^x\) |
| \(a^x\) | \(a^x \ln a\) |
| \(\ln x\) | \(\frac{1}{x}\) |
| \(\log_a x\) | \(\frac{1}{x \ln a}\) |
Additional Information: Differentiation Techniques
Understanding different differentiation techniques is crucial for solving calculus problems. Here are a few related concepts:
- Chain Rule: Used to find the derivative of a composite function. If \(y = f(g(x))\), then \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\). The method used in the problem, \(\frac{dy}{du} = \frac{dy/dx}{du/dx}\), is a direct consequence of the chain rule where \(y\) is a function of \(u\), and \(u\) is a function of \(x\). If \(y = F(u)\) and \(u = g(x)\), then \(y = F(g(x))\). By the chain rule, \(\frac{dy}{dx} = F'(g(x)) \cdot g'(x) = \frac{dF}{du} \cdot \frac{du}{dx}\). Rearranging gives \(\frac{dF}{du} = \frac{dy/dx}{du/dx}\), which is \(\frac{dy}{du} = \frac{dy/dx}{du/dx}\).
- Product Rule: Used to find the derivative of a product of two functions. If \(h(x) = f(x)g(x)\), then \(h'(x) = f'(x)g(x) + f(x)g'(x)\).
- Quotient Rule: Used to find the derivative of a quotient of two functions. If \(h(x) = \frac{f(x)}{g(x)}\), then \(h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}\).
- Implicit Differentiation: Used to find the derivative of a function defined implicitly by an equation, where \(y\) is not explicitly expressed as a function of \(x\).
Mastering these techniques and fundamental formulas is essential for tackling various differentiation problems in calculus.
Similar Questions
If \({\rm{y}} = {\cos ^{ - 1}}\left( {\frac{{2{\rm{x}}}}{{1 + {{\rm{x}}^2}}}} \right)\) , then \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}}\) is equal to
What is the derivative of log 10 (5x 2+ 3) with respect to x?
The derivative of In(x + sin x) with respect to (x + cos x) is
If y = \(\frac{x \sqrt{x^2−16}}{2} − 8 \ln\left|x + \sqrt{x^2−16}\right|\) , then what is \(\frac{\text{dy}}{\text{dx}}\) equal to ?
If e θϕ = c + 4θϕ, where c is an arbitrary constant and ϕ is a function of θ, then what is ϕ dθ equal to?
If \(\rm x^m y^n =a^{m+n}\) , then what is \(\rm \dfrac{dy}{dx}\) equal to?
What is the derivative of sec 2(tan -1 x) with respect to x?
A function is defined in (0, ∞) by \( f(x) = \begin{cases} 1-x^2 & for& , 0 < x \leq 1 \quad \\ In \ x & for &, 1 < x \leq 2 \\ In \ 2 - 1 + 0.5x & for &, 2 < x < \infty \end{cases} \)
Which one of the following is correct in respect of the derivative of the function, i.e. f’(x)?Let f(x + y) = f(x) f(y) for all x and y. Then what is f’(5) equal to [where f’(x) is the derivative of f(x)]?
Which of the following statements are not correct?
1. y as a function of x is not defined for all real x.
2. y as a function of x is not continuous at x = 0
3. y as a function of x is differentiable for all x.
Select the correct answer using the code given below
Important Questions from Evaluation of derivatives
Differentiate {-log (log x), x > 1} with respect to x
Find the derivation of f(x) = 1/x2
- The derivative of the function f(x) = -3x2 + 6x - 4 is given by:
Differential coefficient of log10 x with respect to logx 10 is
The derivatives of (x3 + ex + 3x + cot x) with respect to x is
