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

Question

If $y=e^{x^{e^x}}+x^{e^{e^x}}+e^{x^{x^e}}$, prove that $$\frac{dy}{dx}=e^{x^{e^x}}x^{e^x}\left({\frac{e^x}x+e^x\log x}\right) +x^{e^{e^x}}e^{e^x}\left({\frac1x+e^x\log x}\right) +e^{x^{x^e}}x^{x^e}x^{e-1}({1+e\log x})$$

Edit:- please check is my approach correct.

Answer 1

HINTS:

1. Differentiate the sum term by term $$\frac{\text{d}}{\text{d}x}\left(f(x)+y(x)\right)=\frac{\text{d}}{\text{d}x}\left(f(x)\right)+\frac{\text{d}}{\text{d}x}\left(y(x)\right)=f'(x)+y'(x)$$
2. When $\text{k}_1$ and $\text{k}_2$ are constants, use the chain rule: $$\frac{\text{d}}{\text{d}x}\left(\text{k}_1^{q(x)^{\text{k}_2^{z(x)}}}\right)=\text{k}_1^{q(x)^{\text{k}_2^{z(x)}}}\cdot\ln\left(\text{k}_1\right)\cdot\frac{\text{d}}{\text{d}x}\left(q(x)^{\text{k}_2^{z(x)}}\right)$$
3. When $\text{k}_2$ is a constant, use the chain rule: $$\frac{\text{d}}{\text{d}x}\left(q(x)^{\text{k}_2^{z(x)}}\right)=\frac{\text{d}}{\text{d}x}\left(e^{\text{k}_2^{z(x)}\ln\left(q(x)\right)}\right)=e^{\text{k}_2^{z(x)}}\cdot\frac{\text{d}}{\text{d}x}\left(\text{k}_2^{z(x)}\ln\left(q(x)\right)\right)$$
4. When $\text{k}_2$ is a constant, use the product rule: $$\frac{\text{d}}{\text{d}x}\left(\text{k}_2^{z(x)}\ln\left(q(x)\right)\right)=\ln\left(q(x)\right)\cdot\frac{\text{d}}{\text{d}x}\left(\text{k}_2^{z(x)}\right)+\text{k}_2^{z(x)}\cdot\frac{\text{d}}{\text{d}x}\left(\ln\left(q(x)\right)\right)$$
5. When $\text{k}_2$ is a constant, use the chain rule: $$\frac{\text{d}}{\text{d}x}\left(\text{k}_2^{z(x)}\right)=\text{k}_2^{z(x)}\ln\left(\text{k}_2\right)\cdot\frac{\text{d}}{\text{d}x}\left(z(x)\right)=\text{k}_2^{z(x)}\ln\left(\text{k}_2\right)z'(x)$$
6. Use the chain rule: $$\frac{\text{d}}{\text{d}x}\left(\ln\left(q(x)\right)\right)=\frac{\frac{\text{d}}{\text{d}x}\left(q(x)\right)}{q(x)}=\frac{q'(x)}{q(x)}$$