Question

How do you implicitly differentiate `11=(x)/(ye^x)`?

Answer 1

`dy/dx=(1 -11ye^x)/(11e^x)`

Explanation:

`11=x/(ye^x)`
can be rearranged as;
`11ye^x=x`

We can now differentiate wrt x;

`d/dx{(11y)(e^x)} = d/dx(x)`

We can use the product rule to differentiate the LHS and we know already know how to differentiate the RHS:

So, `d/dx(11ye^x)=d/dx(x)`
`:. (11y)(d/dxe^x)+(d/dx11y)(e^x)=d/dx(x)`
`:. (11y)e^x+(d/dx11y)(e^x)=1`

We can then use the chain rule as `(dA)/dx = (dA)/dy.dy/dx`; so
`11ye^x+(d/dy11y)dy/dx(e^x)=1`
`11ye^x+11dy/dxe^x=1`
`11e^xdy/dx=1 -11ye^x`
Hence, `dy/dx=(1 -11ye^x)/(11e^x)`