Differentiate $[x^{5\coth(6x)}]'$

Question

can you help me to differentiate this function? $$[x^{5\coth(6x)}]'$$

My steps:

$$[x^{5\coth(6x)}*\ln(x)]*[5(1-\coth^2(6x))]*[6]$$

I dont know what formula i should use $$[x^n]'$$ or $$[a^x]'$$ thanks for advice.

Answer 1

HINT:

$$\frac{\text{d}}{\text{d}x}\left(x^{5\coth(6x)}\right)=$$ $$\frac{\text{d}}{\text{d}x}\left(e^{\left(5\coth(6x)\right)\ln(x)}\right)=$$ $$\left(\frac{\text{d}}{\text{d}x}\left(5\coth(6x)\ln(x)\right)\right)e^{\left(5\coth(6x)\right)\ln(x)}=$$ $$x^{5\coth(6x)}\cdot\frac{\text{d}}{\text{d}x}(5\coth(6x)\ln(x))=$$ $$5x^{5\coth(6x)}\cdot\frac{\text{d}}{\text{d}x}(\coth(6x)\ln(x))=$$ $$5x^{5\coth(6x)}\cdot\left(\ln(x)\frac{\text{d}}{\text{d}x}(\coth(6x))+\coth(6x)\frac{\text{d}}{\text{d}x}(\ln(x))\right)=$$ $$5x^{5\coth(6x)}\cdot\left(\ln(x)\frac{\text{d}}{\text{d}x}(\coth(6x))+\coth(6x)\cdot\frac{1}{x}\right)$$

Answer 2

Hint:

$$y(x)=x^{f(x)}\Leftrightarrow \ln(y(x))=f(x)\ln(x)$$

so that $$\frac{y'(x)}{y(x)}=\left(f(x)\ln(x)\right)'$$