1. **Stating the problem:** We are asked to find the derivative with respect to $x$ and $y$ of a very complex nested function involving trigonometric, logarithmic, exponential, and power functions. 2. **Understanding the problem:** The expression involves multiple layers of functions such as $\sin^e(\cos^n(\tan^s(\ln e + y^s) + (\ln y)^e e^{q^h} + 3(y^s e)^e))$, powers of constants, logarithms, and compositions of trigonometric and inverse trigonometric functions. 3. **Key formulas and rules:** - Derivative of $\sin u$ is $\cos u \cdot \frac{du}{dx}$. - Derivative of $\cos u$ is $-\sin u \cdot \frac{du}{dx}$. - Derivative of $\tan u$ is $\sec^2 u \cdot \frac{du}{dx}$. - Derivative of $\ln u$ is $\frac{1}{u} \cdot \frac{du}{dx}$. - Chain rule: $\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$. - Power rule: $\frac{d}{dx} u^n = n u^{n-1} \cdot \frac{du}{dx}$. 4. **Simplifications:** - $\ln e = 1$. - Constants like $e$, $\pi$, and numbers are treated as constants. 5. **Stepwise differentiation:** Since the expression is very complex and involves both $x$ and $y$, we focus on differentiating with respect to $x$ first, then $y$. **Example for a part:** Consider $f(y) = \sin^e(\cos^n(\tan^s(\ln e + y^s) + (\ln y)^e e^{q^h} + 3(y^s e)^e))$. - Let $u = \cos^n(\tan^s(1 + y^s) + (\ln y)^e e^{q^h} + 3(y^s e)^e)$. - Then $f(y) = (\sin u)^e$. Derivative w.r.t $y$: $$ \frac{df}{dy} = e (\sin u)^{e-1} \cos u \cdot \frac{du}{dy} $$ - Next, differentiate $u = (\cos v)^n$ where $v = \tan^s(1 + y^s) + (\ln y)^e e^{q^h} + 3(y^s e)^e$. $$ \frac{du}{dy} = n (\cos v)^{n-1} (-\sin v) \cdot \frac{dv}{dy} $$ - Then differentiate $v$ term by term using chain and product rules. 6. **Final answer:** Due to the extreme complexity and nested nature, the derivative is expressed as a combination of chain rule applications on each nested function, carefully applying power, product, and chain rules. **Summary:** The derivative involves repeated application of the chain rule on nested functions: $$ \frac{d}{dy} \sin^e(\cos^n(\tan^s(1 + y^s) + (\ln y)^e e^{q^h} + 3(y^s e)^e)) = e (\sin u)^{e-1} \cos u \cdot n (\cos v)^{n-1} (-\sin v) \cdot \frac{dv}{dy} $$ where $$ \frac{dv}{dy} = s \tan^{s-1}(1 + y^s) \sec^2(1 + y^s) s y^{s-1} + e (\ln y)^{e-1} \frac{1}{y} e^{q^h} + 3 e (y^s e)^{e-1} s y^{s-1} e $$ and so forth for other parts. This approach applies similarly for derivatives w.r.t $x$ and other terms. --- "slug":"complex derivative", "subject":"calculus", "desmos":{ "latex":"y=\sin^e(\cos^n(\tan^s(\ln e + y^s) + (\ln y)^e e^{q^h} + 3(y^s e)^e))", "features":{ "intercepts":true, "extrema":true } }, "q_count":1