What is the derivative of \(\rm e^{e^x}\) with respect to e x?
The question asks for the derivative of the function \(y = \rm e^{e^x}\) with respect to \(e^x\). This is different from finding the derivative with respect to \(x\). To solve this, we can use a substitution method.
Let's simplify the problem by introducing a new variable.
We are asked to find the derivative of \(y\) with respect to \(u\), which is denoted as \(\frac{dy}{du}\).
To find \(\frac{dy}{du}\), we differentiate \(y = e^u\) with respect to \(u\).
The derivative of the exponential function \(e^z\) with respect to \(z\) is \(e^z\).
So, \(\frac{dy}{du} = \frac{d}{du}(e^u) = e^u\).
Now we substitute back the original expression for \(u\), which is \(u = e^x\).
So, \(\frac{dy}{du} = e^u = e^{e^x}\).
Thus, the derivative of \(\rm e^{e^x}\) with respect to \(\rm e^x\) is \(\rm e^{e^x}\).
Let's look at the given options:
Our calculated derivative, \( \rm e^{e^x} \), matches option 1.
Note that option 3, \( \rm e^{e^x} e^x \), would be the derivative of \( \rm e^{e^x} \) with respect to \(x\), which is found using the chain rule: \(\frac{d}{dx}(e^{e^x}) = e^{e^x} \cdot \frac{d}{dx}(e^x) = e^{e^x} \cdot e^x\).
Therefore, the correct answer is \( \rm e^{e^x} \).
| Function | Derivative with respect to \(x\) | Derivative with respect to \(u = f(x)\) |
|---|---|---|
| \(e^x\) | \(e^x\) | N/A (usually differentiate w.r.t x) |
| \(e^u\) | \(e^u \cdot \frac{du}{dx}\) (Chain Rule) | \(e^u\) |
| \(e^{e^x}\) | \(e^{e^x} \cdot e^x\) (Derivative w.r.t. \(x\)) | \(e^{e^x}\) (Derivative w.r.t. \(e^x\)) |
| Function | Derivative |
|---|---|
| \(c\) (constant) | \(0\) |
| \(x^n\) | \(nx^{n-1}\) |
| \(e^x\) | \(e^x\) |
| \(\ln|x|\) | \(\frac{1}{x}\) |
| \(a^x\) | \(a^x \ln a\) |
| \(\sin x\) | \(\cos x\) |
| \(\cos x\) | \(-\sin x\) |
| Chain Rule: \(f(g(x))\) | \(f'(g(x)) \cdot g'(x)\) |
| Product Rule: \(u(x)v(x)\) | \(u'(x)v(x) + u(x)v'(x)\) |
| Quotient Rule: \(\frac{u(x)}{v(x)}\) | \(\frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\) |
Differentiation is a fundamental concept in calculus that measures the rate at which a function changes with respect to its input variable. The derivative of a function \(f(x)\) with respect to \(x\) is denoted as \(f'(x)\) or \(\frac{df}{dx}\).
When asked for the derivative of a function \(y\) with respect to another function of the same variable, say \(u(x)\), you are essentially looking for \(\frac{dy}{du}\). If \(y = f(u)\), then \(\frac{dy}{du} = f'(u)\).
This can be understood using the chain rule. We know that \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\). If \(\frac{du}{dx} \neq 0\), we can write \(\frac{dy}{du} = \frac{dy/dx}{du/dx}\). In our case, \(y = e^{e^x}\) and \(u = e^x\). \(\frac{dy}{dx} = e^{e^x} \cdot e^x\) \(\frac{du}{dx} = e^x\) So, \(\frac{dy}{du} = \frac{e^{e^x} \cdot e^x}{e^x} = e^{e^x}\). This confirms the result obtained through direct substitution.
The number \(e\) is a special mathematical constant approximately equal to 2.71828. The function \(f(x) = e^x\) is the unique function that is equal to its own derivative, i.e., \(\frac{d}{dx}(e^x) = e^x\).
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