How do you find the first and second derivative of $y = x^{e^{c} x}$ $y=x^(e^cx)$ ?

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Answer 1 · 2017-08-10T06:59:33

$(1) : \frac{d y}{d x} = e^{c} y (1 + ln x) = e^{c} x^{e^{c} \cdot x} (1 + ln x) .$ $(1) : dy/dx=e^cy(1+lnx)=e^cx^(e^c*x)(1+lnx).$

$(2) : \frac{d^{2} y}{{d x}^{2}} = e^{c} x^{e^{c} \cdot x - 1} {1 + x e^{c} {(1 + ln x)}^{2}} .$ $(2) : (d^2y)/dx^2=e^cx^(e^c*x-1){1+xe^c(1+lnx)^2}.$

$y = x^{e^{c} \cdot x}$ $y=x^(e^c*x)$

$:. ln y=ln{x^(e^c*x)}=(e^c*x)(lnx).$

$:. d/dx{lny}=d/dx{(e^c*x)(lnx)}.$

Applying the Chain Rule for the L.H.S., and, the

Product Rule for the R.H.S., we get,

${d/dy(lny)}{dy/dx}=(xe^c)d/dx(lnx)+(lnx)d/dx((xe^c)).$

$:. 1/ydy/dx=(xe^c)(1/x)+(lnx)(e^cd/dx(x)).$

$=e^c+(e^c)(lnx)(1)=e^c(1+lnx)$

$rArr dy/dx=e^cy(1+lnx)..............(d_1).$

Next, $(d^2y)/dx^2=d/dx{dy/dx},$

$=d/dx{e^cy(1+lnx)},$

$=e^cd/dx{y(1+lnx)},$

$=e^c[yd/dx(1+lnx)+(1+lnx)d/dx(y)],$

$=e^c{y*1/x+(1+lnx)e^cy(1+lnx)}....[because, (d_1)],$

$=y/x*e^c{1+xe^c(1+lnx)^2}.$

Since, $y=x^(e^c*x),$ we have,

$(d^2y)/dx^2=e^cx^(e^c*x-1){1+xe^c(1+lnx)^2}.$

Enjoy Maths.!

How do you find the first and second derivative of y=x^(e^cx)y=xecxy=x^(e^cx)?