Question

Find y' in these 2 equations?

`x.e^(-y/2)` `+ y.e^(-x/2)` `=sin(x^2 + y^2)`
`and`

`y=x^(lnx) . (sec x)^(3x)`

Answer 1

`y'=(4xcos(x^2+y^2)-2e^(-y/2)+ye^(-x/2))/(2e^(-x/2)-xe^(-y/2)-4ycos(x^2+y^2))`

Explanation:

Let , `y'=dy/dx=y_1`

Here ,

`(1)x*e^(-y/2)+y*e^(-x/2)=sin(x^2+y^2)`

Diff. w.r.t. `x`,using product and chain rules :

`xe^(-y/2)(-1/2)y_1+e^(-y/2)+ye^(-x/2)(-1/2)+e^(-x/2) *y_1`

`color(white)(......................................)=cos(x^2+y^2)[2x+2yy_1]`

`:.xe^(-y/2)(-1/2)y_1+e^(-x/2) *y_1-cos(x^2+y^2)2yy_1`

`color(white)(...........................)=2xcos(x^2+y^2)-e^(-y/2)-ye^(-x/2)(-1/2)`

`y_1{e^(-x/2)-x/2e^(-y/2)-2ycos(x^2+y^2)}`

`color(white)(...........................)=2xcos(x^2+y^2)-e^(-y/2)+y/2e^(-x/2)`

`y_1{2e^(-x/2)-xe^(-y/2)-4ycos(x^2+y^2)}`

`color(white)(...........................)=4xcos(x^2+y^2)-2e^(-y/2)+ye^(-x/2)`

`:.y_1=(4xcos(x^2+y^2)-2e^(-y/2)+ye^(-x/2))/(2e^(-x/2)-xe^(-y/2)-4ycos(x^2+y^2))`

For second eqn. ,please see next answer.

Answer 2

`(dy)/(dx)=x^(2lnx-1)*lnx^2+3(secx)^(6x)[xtanx+ln(secx)]`

Explanation:

Let , `y'=(dy)/(dx)=y_1`

`(2)y=x^lnx*(secx)^(3x)=u*v.......................(i)`

Let ,

`u=x^lnx` ,`..................and v=(secx)^(3x)`

Taking natural log both sides we get

`lnu=ln(x^lnx)` , `................andlnv=ln((secx)^(3x))`

`:.lnu=lnx*lnx =(lnx)^2 ............and lnv=3xln(secx)`

Diff.w.r.t. `x`

`1/u(du)/(dx)=2lnx*1/xand 1/v(dv)/(dx)`=`3x*(secxtanx)/secx+ln(secx)*3`

`1/u(du)/(dx)=1/xlnx^2............and 1/v(dv)/(dx)`=`3xtanx+3ln(secx)`

`(du)/(dx)=u/xlnx^2............and (dv)/(dx)=3v[xtanx+ln(secx)]....(ii)`

Now ,

`y=u*vto "Product Rule"`

`(dy)/(dx)=u(dv)/(dx)+v(du)/(dx)`

From `(i) and (ii)`

`:.(dy)/(dx)=x^lnx [u/xlnx^2]+(secx)^(3x)[3v(xtanx+ln(secx))]`

Subst. back values of `u and v`
`(dy)/(dx)`=`x^lnx [x^lnx/xlnx^2]+(secx)^(3x)[3(secx)^(3x)(xtanx+ln(secx))]`

`(dy)/(dx)=x^(2lnx-1)*lnx^2+3(secx)^(6x)[xtanx+ln(secx)]`