Question

If H= tan^-1(x/y) x=u+v and y=u-v then dH/dv ?

Answer 1

`(dH)/(dv) = -u/(u^2+v^2)`

Explanation:

Given: `H= tan^-1(y/x)`

Substitute `x=u+v` and `y = u-v`:

`H= tan^-1((u-v)/(u+v))`

Let `g= (u-v)/(u+v)`, then use the chain rule:

`(dH)/(dv) = (d(tan^-1(g)))/(dg)(dg)/(dv)`

`(d(tan^-1(g)))/(dg) = 1/(g^2+1)= 1/(((u-v)/(u+v))^2+1) = (u+v)^2/((u-v)^2+(u+v)^2)`

Use the quotient rule for `g= (u-v)/(u+v)`:

`(dg)/(dv) = ((d(u-v))/(dv)(u+v)-(d(u+v))/(dv)(u-v))/(u+v)^2`

Please notice that we must treat `u` as a constant because we do not have any information that would allow us to implicitly differentiate it.

`(dg)/(dv) = ((-1)(u+v)-(1)(u-v))/(u+v)^2`

`(dg)/(dv) = (-u-v-u+v)/(u+v)^2`

`(dg)/(dv) = (-2u)/(u+v)^2`

Substitute into the chain rule:

`(dH)/(dv) = (u+v)^2/((u-v)^2+(u+v)^2)(-2u)/(u+v)^2`

`(dH)/(dv) = (-2u)/((u-v)^2+(u+v)^2)`

`(dH)/(dv) = (-2u)/((u^2-2uv+v^2+u^2+2uv+v^2)`

`(dH)/(dv) = -u/(u^2+v^2)`

Answer 2

`(dH)/(dv)=u/(u^2+v^2)`

Explanation:

`H=tan^-1((u+v)/(u-v))`

`"differentiate using the "color(blue)"chain rule"`

`"given "y=f(g(x))" then"`

`dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"`

`rArr(dH)/(dv)=1/(1+((u+v)/(u-v)))^2xxd/(dv)((u+v)/(u-v))`

`"differentiate "((u+v)/(u-v))" using the "color(blue)"quotient rule"`

`"given "y=(g(x))/(h(x))" then"`

`dy/dx=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larrcolor(blue)"quotient rule"`

`g(v)=u+vrArrg'(v)=1`

`h(v)=u-vrArrh'(v)=-1`

`rArrd/(dv)((u+v)/(u-v))=(u-v+u+v)/(u-v)^2=(2u)/(u-v)^2`

`rArr(dH)/(dv)=1/(1+(u+v)^2/(u-v)^2)xx(2u)/(u-v)^2`

`color(white)(xxxxx)=(2u)/((u-v)^2+(u+v)^2)`

`color(white)(xxxxx)=(2u)/(u^2cancel(-2uv)+v^2+u^2cancel(+2uv)+v^2)`

`color(white)(xxxx)=(2u)/(2u^2+2v^2)=u/(u^2+v^2)`