Question

Find the cosine of angle of intersection of curves $y=2^x\ell nxandy=x^{2x}-1at\left(1,0\right)$

Answer 1

$y=2^x\ln x,y=x^{2x}-1$ at $(1,0)$

angle between the intersection of urve is angle between their tangents at point of intersection.

$\therefore \frac{dy}{dx}=\frac{d\left(2^x\ln x\right)}{dx}=\frac{2^x}{x}+2^x\ln 2.\ln x\Rightarrow {\left(\frac{dy}{dx}\right)}_{(1,0)}=2=m_1$

Similarly for another curve,

$\frac{dy}{dx}=\frac{d(x^{2x}-1)}{dx}={2x}^{2x}(1+\ln x){\left(\frac{dy}{dx}\right)}_{(1,0)}=2=m_2$

$\therefore$ angle between the curves be $\theta$

$\therefore \tan \theta =\frac{m_2-m_1}{1+m_1{m}_2}=0\Rightarrow \theta =0\Rightarrow \cos \theta =1$