Question #88084

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2 Answers
Pankaj Solanki
Sep 17, 2017

#(dy)/dx=(e^x(x-2))/x^3#

Explanation:

#y = (e^x)/(x^2)-1#
using differential formula
#d/dx (u/v)=((du)/dx.v-(dv)/dx.u)/v^2#

hence here #u=e^x# and #v=x^2#
#(dy)/dx=((d/dx(e^x)).x^2-(d/dx(x^2)).e^x)/(x^2)^2-d/dx(1)#
#(dy)/dx=((e^x).x^2-(2x).e^x)/(x^4)#
#(dy)/dx=(e^x.x^2-2x.e^x)/(x^4)=(xe^x-2e^x)/x^3#
#(dy)/dx=(e^x(x-2))/x^3#

Tazwar Sikder
Sep 17, 2017

#y' = frac(e^(x) (x - 2))(x^(3))#

Explanation:

We have: #y = frac(e^(x))(x^(2)) - 1#

#Rightarrow y' = frac(d)(dx) (frac(e^(x))(x^(2))) - frac(d)(dx) (1)#

#Rightarrow y' = frac(d)(dx) (frac(e^(x))(x^(2))) - 0#

#Rightarrow y' = frac(d)(dx) (frac(e^(x))(x^(2)))#

#Rightarrow y' = frac(x^(2) cdot frac(d)(dx) (e^(x)) - e^(x) cdot frac(d)(dx) (x^(2)))((x^(2))^(2))#

#Rightarrow y' = frac(x^(2) cdot e^(x) - e^(x) cdot 2 x)(x^(4))#

#Rightarrow y' = frac(e^(x) (x^(2) - 2 x))(x^(4))#

#Rightarrow y' = frac(x e^(x) (x - 2))(x^(4))#

#therefore y' = frac(e^(x) (x - 2))(x^(3))#

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