<p>Show that</p>
<ol class="abc-list"><li><div class="list-item-paragraph">$\frac{\mathrm{d}}{\mathrm{d}x}{{\sin }^{-1}}x=\frac{1}{\sqrt{1-{{x}^{2}}}}$</div></li><li><div class="list-item-paragraph">$\frac{\mathrm{d}}{\mathrm{d}x}{{\tan }^{-1}}x=\frac{1}{1+{{x}^{2}}}$</div></li><li><div class="list-item-paragraph">$\frac{\mathrm{d}}{\mathrm{d}x}{{\cos }^{-1}}x=-\frac{1}{\sqrt{1-{{x}^{2}}}}$</div></li></ol>

$\frac{\mathrm{d}}{\mathrm{d}x}{{\sin }^{-1}}x=\frac{1}{\sqrt{1-{{x}^{2}}}}$

$\frac{\mathrm{d}}{\mathrm{d}x}{{\tan }^{-1}}x=\frac{1}{1+{{x}^{2}}}$

$\frac{\mathrm{d}}{\mathrm{d}x}{{\cos }^{-1}}x=-\frac{1}{\sqrt{1-{{x}^{2}}}}$

Techniques of Differentiation - General

Techniques of Differentiation

H2 Math

International Baccalaureate Higher Level Mathematics

Techniques of Differentiation - Techniques of Differentiation - General

Question 34b93b91

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