If y=enx, then d2ydx2d2xdy2 is equal to

A
${ne}^{nx}$
B
${ne}^{- nx}$
C
1
D
$- {ne}^{- nx}$

Solution:

The given function is $y = e^{n x}$

Differentiate both sides

$\frac{d y}{d x} = n e^{n x}$

Again differentiate both sides

$\frac{d^{2} y}{d x^{2}} = n^{2} e^{n x} n o w y = e^{n x} \Rightarrow n x = \log y \Rightarrow n \frac{d x}{d y} = \frac{1}{y} \Rightarrow n \frac{d^{2} x}{d y^{2}} = \frac{- 1}{y^{2}} \Rightarrow \frac{d^{2} x}{d y^{2}} = \frac{- 1}{n y^{2}} \Rightarrow \frac{d^{2} y}{d x^{2}} \frac{d^{2} x}{d y^{2}} = n^{2} y \times \frac{- 1}{n y^{2}} = - \frac{n}{y} = - n e^{- n x}$

If $y = e^{nx},$ then $(\frac{d^{2} y}{{dx}^{2}}) (\frac{d^{2} x}{{dy}^{2}})$ is equal to

Solution:

Related content