The Notes: Find the derivative of xsinx by first principle.

Monday, June 6, 2011

Find the derivative of xsinx by first principle.

Hello!

Given a function $\displaystyle{f{{\left({x}\right)}}}={x}{\sin{{\left({x}\right)}}},$ and we have to find its derivative by the definition. Consider the expression $\displaystyle\frac{{{f{{\left({x}+\delta{x}\right)}}}-{f{{\left({x}\right)}}}}}{{\delta{x}}}$ and find its limit for $\displaystyle\delta{x}\to{0}:$

$\displaystyle\frac{{{f{{\left({x}+\delta{x}\right)}}}-{f{{\left({x}\right)}}}}}{{\delta{x}}}=\frac{{{\left({x}+\delta{x}\right)}{\sin{{\left({x}+\delta{x}\right)}}}-{x}{\sin{{\left({x}\right)}}}}}{{\delta{x}}}=$

$\displaystyle=\frac{{{x}{\left({\sin{{\left({x}+\delta{x}\right)}}}-{\sin{{\left({x}\right)}}}\right)}+\delta{x}{\sin{{\left({x}+\delta{x}\right)}}}}}{{\delta{x}}}=$

$\displaystyle={x}\frac{{{\sin{{\left({x}+\delta{x}\right)}}}-{\sin{{\left({x}\right)}}}}}{{\delta{x}}}+{\sin{{\left({x}+\delta{x}\right)}}}.$

The second summand has the limit $\displaystyle{\sin{{\left({x}\right)}}},$ because $\displaystyle{\sin{{\left({x}\right)}}}$ is continuous for any $\displaystyle{x}.$

The first summand has the limit $\displaystyle{x}{\left({\sin{{\left({x}\right)}}}\right)}'={x}{\cos{{\left({x}\right)}}}$ (by the definition of the derivative, applied to $\displaystyle{\sin{{\left({x}\right)}}}$ ).

So the answer is $\displaystyle{\left({x}{\sin{{\left({x}\right)}}}\right)}'={x}{\cos{{\left({x}\right)}}}+{\sin{{\left({x}\right)}}}.$

If we "don't know" what the derivative of $\displaystyle{\sin{{\left({x}\right)}}}$ is, transform the expression:

$\displaystyle\frac{{{\sin{{\left({x}+\delta{x}\right)}}}-{\sin{{\left({x}\right)}}}}}{{\delta{x}}}=\frac{{{2}{\sin{{\left(\frac{{\delta{x}}}{{2}}\right)}}}{\cos{{\left({x}+\frac{{\delta{x}}}{{2}}\right)}}}}}{{\delta{x}}}=$

$\displaystyle=\frac{{\sin{{\left(\frac{{\delta{x}}}{{2}}\right)}}}}{{\frac{{\delta{x}}}{{2}}}}\cdot{\cos{{\left({x}+\frac{{\delta{x}}}{{2}}\right)}}}.$

The second factor has the limit $\displaystyle{\cos{{\left({x}\right)}}}$ because $\displaystyle{\cos{{\left({x}\right)}}}$ is continuous for any $\displaystyle{x},$ the first has the limit 1 (we should know this).

The Notes

Monday, June 6, 2011

Find the derivative of xsinx by first principle.

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