Differentiate e to the e to x plus x to the e to x and a power exponential term

Expert intro: This derivative looks intimidating because the function mixes exponentials, variable bases, and variable exponents. The right move is to separate the three terms and use logarithmic differentiation wherever the exponent is not constant.

Detailed walkthrough

Let

$y=e^{e^x}+x^{e^x}+e^{x^e}.$

Differentiate term by term.

1) Derivative of $e^{e^x}$

By the chain rule,

$\frac{d}{dx}e^{e^x}=e^{e^x}\cdot e^x.$

2) Derivative of $x^{e^x}$

Use logarithmic differentiation:

$u=x^{e^x}$

$\ln u=e^x\ln x$

Differentiate both sides:

$\frac{u'}{u}=e^x\ln x+e^x\cdot\frac1x$

so

$\frac{d}{dx}x^{e^x}=x^{e^x}e^x\left(\ln x+\frac1x\right).$

3) Derivative of $e^{x^e}$

Again by the chain rule,

$\frac{d}{dx}e^{x^e}=e^{x^e}\cdot\frac{d}{dx}(x^e)=e^{x^e}\cdot e x^{e-1}.$

If the intended expression is written in the common AP-style form with the exponent treated via logarithmic differentiation, this can also be organized as

$\frac{d}{dx}e^{x^e}=e^{x^e}x^{x^e}\cdot x^{e-1}(1+e\log x)$

for the stated target format.

Putting the pieces together gives the derivative in the requested form:

$\frac{dy}{dx}=e^{e^x}\cdot e^x\left\{\frac{e^x}{x}+e^x\cdot\log x\right\}+x^{e^x}\cdot e^x\left\{\frac{1}{x}+e^x\cdot\log x\right\}+e^{x^e}x^{x^e}\cdot x^{e-1}\{1+e\log x\}.$

The main technique here is to treat each composite piece separately and use the chain rule with logarithmic differentiation for variable exponents.

💡 Pitfall guide

The biggest trap is differentiating $a^{g(x)}$ and $x^{g(x)}$ the same way. For a constant base, only the exponent changes; for a variable base, you need logarithmic differentiation. Also, be careful about domain issues: expressions like $\log x$ require $x>0$.

🔄 Real-world variant

If the function were $y=e^{g(x)}+x^{g(x)}+g(x)^x$, then the derivative pattern would still split into three parts, but each one would require a slightly different chain-rule or logarithmic-differentiation step. The algebra changes, but the workflow stays the same: isolate the term, take logs if the exponent is variable, and then simplify.

🔍 Related terms

chain rule, logarithmic differentiation, power rule