Derivative of a complex function $y=\operatorname{tg}2x^{\cot\frac x 2}$

Question

I have: $y=(\operatorname{tg}2x)^{\operatorname{ctg}\frac x 2}$

So I solve: \begin{align} y' & =(\operatorname{tg}2x)^{\operatorname{ctg}\frac x 2} \cdot \ln\operatorname{tg}2x \cdot (\operatorname{ctg}\frac x 2)' \\[10pt] & =(\operatorname{tg}2x)^{\operatorname{ctg}\frac x 2} \cdot \ln\operatorname{tg}2x \cdot -\frac 1 { \sin\frac x 2^2 } \cdot (\frac x 2) ' \\[10pt] & = (\operatorname{tg}2x)^{\operatorname{ctg} \frac x 2} \cdot \ln\operatorname{tg}2x \cdot (-\frac{ 1 }{ \sin^2\frac x 2}) \cdot \frac 1 2 = - \frac{ ({\operatorname{tg}2x})^{\operatorname{ctg}{\frac x 2} } \cdot \ln \operatorname{tg}2x}{2\sin^2{\frac x 2} } \end{align}

But, they told me that I was mistaken. Why so?

Answer 1

$y = \tan^{\cot(\frac x2)}(2x)$

$\ln(y) = \cot(\frac x2)\ln(\tan{(2x)})$

Differentiating wrt x;

$$\frac{y'}{y} = -\dfrac{\operatorname{cosec}^2(\frac x2)}2\cdot\ln(\tan(2x))+\dfrac{1}{\tan(2x)}\cdot\sec^2(2x)\cdot2\cdot\cot(\frac x2)$$

$$y' = y\bigg[-\frac{\operatorname{cosec}^2(\frac x2)} 2 \cdot \ln(\tan(2x)) + 2\cot(2x)\cdot\cot(\frac x2)\bigg]$$

$$y'=\tan^{\cot(\frac x2)}(2x)\bigg[-\dfrac{\operatorname{cosec}^2(\frac x2)}2\cdot\ln(\tan(2x))+2\cot(2x)\cdot\cot(\frac x2)\bigg]$$

Your mistake was when you forgot to differentiate $\ln(\tan(2x))$

generally for functions of the type $y=f(x)^{g(x)}$ use $\ln(y)= g(x) \cdot \ln(f(x))$

Answer 2

Generally you have $$ \frac d {dx} u^v = u^v \log_e u \cdot \frac {dv}{dx} \quad + \quad vu^{v-1} \cdot \frac {du}{dx}. $$ The first term is done just as if $u$ were constant, and the second as if $v$ were constant.