`y = 25arcsin(x/5) -xsqrt(25-x^2)` Find the derivative of the function

Thursday, June 11, 2015

$\displaystyle{y}={25}{\arcsin{{\left(\frac{{x}}{{5}}\right)}}}-{x}\sqrt{{{25}-{{x}}^{{2}}}}$ Find the derivative of the function

$\displaystyle{y}={25}{\arcsin{{\left(\frac{{x}}{{5}}\right)}}}-{x}\sqrt{{{25}-{{x}}^{{2}}}}$

Before taking the derivative, express the radical in exponent form.

$\displaystyle{y}={25}{\arcsin{{\left(\frac{{x}}{{5}}\right)}}}-{x}{{\left({25}-{{x}}^{{2}}\right)}}^{{\frac{{1}}{{2}}}}$

To get y', take the derivative of each term.

$\displaystyle{y}'=\frac{{d}}{{\left.{d}{x}}}{\left[{25}{\arcsin{{\left(\frac{{x}}{{5}}\right)}}}\right]}-\frac{{d}}{{\left.{d}{x}}}{\left[{x}{{\left({25}-{{x}}^{{2}}\right)}}^{{\frac{{1}}{{2}}}}\right]}$

$\displaystyle{y}'={25}\frac{{d}}{{\left.{d}{x}}}{\left[{\arcsin{{\left(\frac{{x}}{{5}}\right)}}}\right]}-\frac{{d}}{{\left.{d}{x}}}{\left[{x}{{\left({25}-{{x}}^{{2}}\right)}}^{{\frac{{1}}{{2}}}}\right]}$

Take note that the derivative formula of arcsine is $\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left[{\arcsin{{\left({u}\right)}}}\right]}=\frac{{1}}{\sqrt{{{1}-{{u}}^{{2}}}}}\cdot\frac{{{d}{u}}}{{\left.{d}{x}}}$ .

Applying that formula, y' will become:

$\displaystyle{y}'={25}\cdot\frac{{1}}{\sqrt{{{1}-{{\left(\frac{{x}}{{5}}\right)}}^{{2}}}}}\cdot\frac{{d}}{{\left.{d}{x}}}{\left(\frac{{x}}{{5}}\right)}-\frac{{d}}{{\left.{d}{x}}}{\left[{x}{{\left({25}-{{x}}^{{2}}\right)}}^{{\frac{{1}}{{2}}}}\right]}$

$\displaystyle{y}'={25}\cdot\frac{{1}}{\sqrt{{{1}-{{\left(\frac{{x}}{{5}}\right)}}^{{2}}}}}\cdot\frac{{1}}{{5}}-\frac{{d}}{{\left.{d}{x}}}{\left[{x}{{\left({25}-{{x}}^{{2}}\right)}}^{{\frac{{1}}{{2}}}}\right]}$

$\displaystyle{y}'={25}\cdot\frac{{1}}{\sqrt{{{1}-\frac{{{x}}^{{2}}}{{25}}}}}\cdot\frac{{1}}{{5}}-\frac{{d}}{{\left.{d}{x}}}{\left[{x}{{\left({25}-{{x}}^{{2}}\right)}}^{{\frac{{1}}{{2}}}}\right]}$

$\displaystyle{y}'={25}\cdot\frac{{1}}{{{\left(\frac{{1}}{{5}}\right)}\sqrt{{{25}-{{x}}^{{2}}}}}}\cdot\frac{{1}}{{5}}-\frac{{d}}{{\left.{d}{x}}}{\left[{x}{{\left({25}-{{x}}^{{2}}\right)}}^{{\frac{{1}}{{2}}}}\right]}$

$\displaystyle{y}'=\frac{{25}}{\sqrt{{{25}-{{x}}^{{2}}}}}-\frac{{d}}{{\left.{d}{x}}}{\left[{x}{{\left({25}-{{x}}^{{2}}\right)}}^{{\frac{{1}}{{2}}}}\right]}$

To take the derivative of the second term, apply the product rule $\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left({u}\cdot{v}\right)}={u}\cdot\frac{{{d}{v}}}{{\left.{d}{x}}}+{v}\cdot\frac{{{d}{u}}}{{\left.{d}{x}}}$ .

Applying this, the y' will be:

$\displaystyle{y}'=\frac{{25}}{\sqrt{{{25}-{{x}}^{{2}}}}}-{\left[{x}\cdot\frac{{d}}{{\left.{d}{x}}}{\left({{\left({25}-{{x}}^{{2}}\right)}}^{{\frac{{1}}{{2}}}}\right)}+{{\left({25}-{{x}}^{{2}}\right)}}^{{\frac{{1}}{{2}}}}\cdot\frac{{d}}{{\left.{d}{x}}}{\left({x}\right)}\right]}$

Also, use the derivative formula $\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left({{u}}^{{n}}\right)}={n}\cdot{{u}}^{{{n}-{1}}}\cdot\frac{{{d}{u}}}{{\left.{d}{x}}}$ .

$\displaystyle{y}'=\frac{{25}}{\sqrt{{{25}-{{x}}^{{2}}}}}-{\left[{x}\cdot\frac{{1}}{{2}}\cdot{{\left({25}-{{x}}^{{2}}\right)}}^{{-\frac{{1}}{{2}}}}\cdot\frac{{d}}{{\left.{d}{x}}}{\left({25}-{{x}}^{{2}}\right)}+{{\left({25}-{{x}}^{{2}}\right)}}^{{\frac{{1}}{{2}}}}\cdot{1}\right]}$

$\displaystyle{y}'=\frac{{25}}{\sqrt{{{25}-{{x}}^{{2}}}}}-{\left[{x}\cdot\frac{{1}}{{2}}\cdot{{\left({25}-{{x}}^{{2}}\right)}}^{{-\frac{{1}}{{2}}}}\cdot{\left(-{2}{x}\right)}+{{\left({25}-{{x}}^{{2}}\right)}}^{{\frac{{1}}{{2}}}}\cdot{1}\right]}$

$\displaystyle{y}'=\frac{{25}}{\sqrt{{{25}-{{x}}^{{2}}}}}-{\left[-{{x}}^{{2}}{{\left({25}-{{x}}^{{2}}\right)}}^{{-\frac{{1}}{{2}}}}+{{\left({25}-{{x}}^{{2}}\right)}}^{{\frac{{1}}{{2}}}}\right]}$

Then, express this with positive exponent only.

$\displaystyle{y}'=\frac{{25}}{\sqrt{{{25}-{{x}}^{{2}}}}}-{\left[-\frac{{{x}}^{{2}}}{{{\left({25}-{{x}}^{{2}}\right)}}^{{\frac{{1}}{{2}}}}}+{{\left({25}-{{x}}^{{2}}\right)}}^{{\frac{{1}}{{2}}}}\right]}$

Also, convert the fractional exponent to radical form.

$\displaystyle{y}'=\frac{{25}}{\sqrt{{{25}-{{x}}^{{2}}}}}-{\left[-\frac{{{x}}^{{2}}}{\sqrt{{{25}-{{x}}^{{2}}}}}+\sqrt{{{25}-{{x}}^{{2}}}}\right]}$

So the derivative of the function simplifies to:

$\displaystyle{y}'=\frac{{25}}{\sqrt{{{25}-{{x}}^{{2}}}}}+\frac{{{x}}^{{2}}}{\sqrt{{{25}-{{x}}^{{2}}}}}-\sqrt{{{25}-{{x}}^{{2}}}}$

$\displaystyle{y}'=\frac{{{{x}}^{{2}}+{25}}}{\sqrt{{{25}-{{x}}^{{2}}}}}-\sqrt{{{25}-{{x}}^{{2}}}}$

$\displaystyle{y}'=\frac{{{{x}}^{{2}}+{25}}}{\sqrt{{{25}-{{x}}^{{2}}}}}-\frac{\sqrt{{{25}-{{x}}^{{2}}}}}{{1}}\cdot\frac{\sqrt{{{25}-{{x}}^{{2}}}}}{\sqrt{{{25}-{{x}}^{{2}}}}}$

$\displaystyle{y}'=\frac{{{{x}}^{{2}}+{25}}}{\sqrt{{{25}-{{x}}^{{2}}}}}-\frac{{{25}-{{x}}^{{2}}}}{\sqrt{{{25}-{{x}}^{{2}}}}}$

$\displaystyle{y}'=\frac{{{{x}}^{{2}}+{25}-{\left({25}-{{x}}^{{2}}\right)}}}{\sqrt{{{25}-{{x}}^{{2}}}}}$

$\displaystyle{y}'=\frac{{{2}{{x}}^{{2}}}}{\sqrt{{{25}-{{x}}^{{2}}}}}$

Therefore, the derivative of the function is $\displaystyle{y}'=\frac{{{2}{{x}}^{{2}}}}{\sqrt{{{25}-{{x}}^{{2}}}}}$ .

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Thursday, June 11, 2015

$\displaystyle{y}={25}{\arcsin{{\left(\frac{{x}}{{5}}\right)}}}-{x}\sqrt{{{25}-{{x}}^{{2}}}}$ Find the derivative of the function

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Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

Thursday, June 11, 2015

Find the derivative of the function

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Post a Comment

Thomas Jefferson&#39;s election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

$\displaystyle{y}={25}{\arcsin{{\left(\frac{{x}}{{5}}\right)}}}-{x}\sqrt{{{25}-{{x}}^{{2}}}}$ Find the derivative of the function

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?