Question

How do you find dy/dt given `y= 4 sin ( sqrt (1+ sqrt (x) ))`?

Answer 1

The chain rule says:

`dy/dt = dy/dx dx/dt`

We can compute `dy/dx`

`dy/dx = 4cos( sqrt (1+ sqrt (x) ))1/(2sqrt (1+ sqrt (x) ))1/(2sqrt(x))`

`dy/dx = cos( sqrt (1+ sqrt (x) ))/(sqrt (1+ sqrt (x) )sqrt(x))`

But, because we are not given a function of x in terms of t, we cannot compute `dx/dt`. Therefore, we must leave the answer in this from:

`dy/dt = cos( sqrt (1+ sqrt (x) ))/(sqrt (1+ sqrt (x) )sqrt(x)) dx/dt`