Question 1

What is the answer to this question?
dy/dx=?

Question 2

$\:\:\:\:\: \:\:\:\:\:\:\star\longrightarrow \sf y = x^(x){}^(²)\\$

Taking the logarithm on both sides -

$\:\:\:\:\: \:\:\:\:\:\:\longrightarrow \sf log y = log x^(x){}^(²)\\$

$\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf log y = x^2 log x\\$

$\:\:\: \boxed{\sf\pink{\:\:\: loga^b = blog a }}\\$

Differentiating with respect to x-

$\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf (d)/(dx) logy = (d)/(dx) x^2 log x \\$

$\:\:\:\:\: \:\:\:\:\:\:\longrightarrow \sf (1)/(y) * (dy)/(dx) = x^2 (d)/(dx) log x + logx (d)/(dx) x^2\\$

$\:\:\:\:\boxed{\sf\pink{(d)/(dx) logx = (1)/(x)}} \\$

$\:\:\:\:\boxed{\sf\pink{\sf(d)/(dx)\bigg[f(x)\:g(x)\bigg] = f(x) (d)/(dx) g(x) + g(x) (d)/(dx) f(x)}}\\$

$\:\:\:\:\: \:\:\:\:\:\:\longrightarrow \sf (d)/(dx) = y \bigg[ x^2 * (1)/(x) + logx * 2x \bigg]\\$

$\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf (dy)/(dx) = y \bigg[ \cancel{x}\: x * \frac{1}{\cancel{x}} + 2x\:logx \bigg]\\$

$\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf \underline{(dy)/(dx) = y \bigg[ x + 2x\:logx \bigg]}\\$

$\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf \underline{(dy)/(dx) = \boxed{\sf x^(x){}^(²)\bigg[ x + 2x\:logx \bigg]}}\\$

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