Secants to tangents: dy/dx

<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text"></span></div><ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Show that $\frac{\mathrm{d}}{\mathrm{d}x}\left( {{3}^{2x}} \right)={{9}^{x}}\ln 9$.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Hence, differentiate ${{\tan }^{-1}}\left( {{3}^{2x}} \right)$, $x>0$ with respect to $x$.</span></div></li></ol></li><li><div class="list-item-content"><span class="list-item-text">Given that ${{y}^{2y}}={{e}^{3x+2e}}$, where $y>{{e}^{-1}}$, show that $y\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}=-\frac{2}{3}{{\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)}^{3}}$.</span></div></li></ol>

Show that $\frac{\mathrm{d}}{\mathrm{d}x}\left( {{3}^{2x}} \right)={{9}^{x}}\ln 9$.

Hence, differentiate ${{\tan }^{-1}}\left( {{3}^{2x}} \right)$, $x>0$ with respect to $x$.

Given that ${{y}^{2y}}={{e}^{3x+2e}}$, where $y>{{e}^{-1}}$, show that $y\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}=-\frac{2}{3}{{\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)}^{3}}$.

2021 SAJC Promo Q5

2021 SAJC Promo Q5

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