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Prabjeet Singh 7 years, 5 months ago
{tex}y=\cfrac {1}{x}{/tex}
{tex}\cfrac {d}{dx} (y)=\cfrac {d}{dx} \bigg( \cfrac{1}{x} \bigg){/tex}
{tex}\cfrac {d}{dx} (y)=\cfrac {d}{dx} (x^{-1}){/tex}
{tex}\cfrac {dy}{dx} = -1(x)^{-1-1}{/tex}
{tex}\cfrac {dy}{dx} = -x^{-2}{/tex}
{tex}\cfrac {dy}{dx} = - \cfrac {1}{x^2}{/tex}
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Prabjeet Singh 7 years, 5 months ago
{tex}{\int \cfrac{(x^2-1)}{(x^2+4)} dx}={\int \cfrac{(x^2+4)-5}{(x^2+4)}dx }{/tex}
{tex}={\int \cfrac{x^2+4}{x^2+4}dx} -5{\int \cfrac {1}{x^2+4}dx}{/tex}
{tex}={\int dx} -5{\int \cfrac {1}{x^2+2^2} dx}{/tex}
{tex}={x-5 \times \cfrac {1}{2}tan^{-1}\cfrac{x}{2}}+c{/tex}
{tex}={x-\cfrac {5}{2} tan^{-1} \cfrac {x}{2} +c}{/tex}
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Jagpreet Singh Bhatia 7 years, 5 months ago
1Thank You