Question

Question #88084

Answer 1

`(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`

Answer 2

`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))`