What is the derivative of this function $y = x sin (\frac{5}{x})$ $y = x sin (5/x)$ ?

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What is the derivative of this function $y = x sin (\frac{5}{x})$ $y = x sin (5/x)$ ?

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Answer 1 · 2017-05-16T11:10:45

$\frac{d y}{d x} = sin (\frac{5}{x}) - \frac{5}{x} cos (\frac{5}{x})$ $dy/dx=sin(5/x)-5/xcos(5/x)$

Explanation:

First we need the product rule, which states that $\frac{d}{d x} (u v) = (\frac{d u}{d x}) v + u (\frac{d v}{d x})$ $d/dx(uv)=((du)/dx)v+u((dv)/dx)$ . Thus:

$\frac{d y}{d x} = (\frac{d}{d x} x) sin (\frac{5}{x}) + x (\frac{d}{d x} sin (\frac{5}{x}))$ $dy/dx=(d/dxx)sin(5/x)+x(d/dxsin(5/x))$

Here, $\frac{d}{d x} x = 1$ $d/dxx=1$ .

To figure out $\frac{d}{d x} sin (\frac{5}{x})$ $d/dxsin(5/x)$ , we need the chain rule since we have a function inside another function. Knowing that $\frac{d}{d x} sin (x) = cos (x)$ $d/dxsin(x)=cos(x)$ , we see that through the chain rule, $\frac{d}{d x} sin (u) = cos (u) (\frac{d u}{d x})$ $d/dxsin(u)=cos(u)((du)/dx)$ .

Then:

$\frac{d y}{d x} = sin (\frac{5}{x}) + x cos (\frac{5}{x}) (\frac{d}{d x} (\frac{5}{x}))$ $dy/dx=sin(5/x)+xcos(5/x)(d/dx(5/x))$

Note that $\frac{d}{d x} \frac{5}{x} = \frac{d}{d x} 5 x^{- 1} = - 5 x^{- 2}$ $d/dx5/x=d/dx5x^-1=-5x^-2$ through the power rule:

$\frac{d y}{d x} = sin (\frac{5}{x}) + x cos (\frac{5}{x}) \cdot (- \frac{5}{x^{2}})$ $dy/dx=sin(5/x)+xcos(5/x)*(-5/x^2)$

$\frac{d y}{d x} = sin (\frac{5}{x}) - \frac{5}{x} cos (\frac{5}{x})$ $dy/dx=sin(5/x)-5/xcos(5/x)$

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What is the derivative of this function $y = x sin (\frac{5}{x})$ $y = x sin (5/x)$ ?

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Explanation:

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Science

Math

Humanities

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What is the derivative of this function y = x sin (5/x)y=xsin(5x)y = x sin (5/x)?

1 Answer

Explanation:

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What is the derivative of this function $y = x sin (\frac{5}{x})$ $y = x sin (5/x)$ ?