Name: Higher Order Derivative
Uploaded: 2019-07-19
Duration: 4 min 20 s
Description: Did you know that functions can be used as exponents? We find derivatives of that function by logarithm. What is a higher-order derivative? Watch this video...

Question

If

y = \sin (\log x)

prove that

x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + y = 0

Accepted Answer

We have, $y = \sin (\log x)$

Now, differentiate $y$ w.r.t $x$ we get

$\Rightarrow \frac{d y}{d x} = \frac{d (\sin (\log x))}{d x}$

$\Rightarrow \frac{d y}{d x} = \cos (\log x) \times \frac{1}{x}$

Now, on multiplying $x$ to both sides we get

$\Rightarrow x \frac{d y}{d x} = \cos (\log x)$ $. . . . . . . . (i)$

Again, differentiate $\frac{d y}{d x}$ w.r.t $x$ we get

$\Rightarrow \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (\frac{\cos (\log x)}{x})$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{- \frac{1}{x} \cdot \sin (\log x) \cdot x - \cos (\log x)}{x^{2}}$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{- \sin (\log x)}{x^{2}} - \frac{\cos (\log x)}{x^{2}}$

$\Rightarrow x^{2} (\frac{d^{2} y}{d x^{2}}) = - \sin (\log x) - \cos (\log x)$ $. . . . . . (ii)$

On adding $(i)$ and $(ii)$ we get

$\Rightarrow x^{2} (\frac{d^{2} y}{d x^{2}}) + x (\frac{d y}{d x}) = - \sin (\log x) - \cos (\log x) + \cos (\log x)$

$\Rightarrow x^{2} (\frac{d^{2} y}{d x^{2}}) + x (\frac{d y}{d x}) = - y [∵ y = \sin (\log x)]$

Hence, $x^{2} (\frac{d^{2} y}{d x^{2}}) + x (\frac{d y}{d x}) + y = 0$

If $y = \sin (\log x)$ prove that $x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + y = 0$

Important Questions on Higher Order Derivatives

Learn and Prepare for any exam you want !

If y=sin(logx) prove that x2d2ydx2+xdydx+y=0

Important Questions on Higher Order Derivatives

Learn and Prepare for any exam you want !

If $y = \sin (\log x)$ prove that $x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + y = 0$