Secants to tangents: dy/dx

<ol class="abc-list"><li><div class="list-item-paragraph">Given that $\sqrt{x}+\sqrt{y}=\sqrt{a}$, where $x>0$, $y>0$ and $a$ is a positive constant, show that</div><div class="list-item-paragraph" style="text-align: center">$2x\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}+\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)-1=0$.</div><p style="text-align: right;">[4]</p></li><li><div class="list-item-paragraph">A curve has parametric equations</div><div class="list-item-paragraph" style="text-align: center">$x=2t-\frac{1}{t}$ and $y=t+\frac{1}{t}$</div><div class="list-item-paragraph">where $t$ is a parameter and $t\ne 0$.</div><div class="list-item-paragraph">Show that $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{2}-\frac{3}{2\left( 2{{t}^{2}}+1 \right)}$.</div><p style="text-align: right;">[2]</p></li></ol>

Given that $\sqrt{x}+\sqrt{y}=\sqrt{a}$, where $x>0$, $y>0$ and $a$ is a positive constant, show that

$2x\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}+\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)-1=0$.

Show that $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{2}-\frac{3}{2\left( 2{{t}^{2}}+1 \right)}$.

2025 RI Techniques of Differentiation Assignment 6A Q2

2025 RI Techniques of Differentiation Assignment 6A Q2

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