Use the Chain Rule to find dw/dt.

w = ln (x^2 + y^2 + z^2)^.5
, x = 9 sin t, y = 6 cos t, z = 7 tan t

1 answer

dw/dt = ∂w/∂x dx/dt + ∂w/∂y dy/dt + ∂w/∂z dz/dt

Using ' to mean d/dt,

w = 1/2 ln(x^2+y^2+z^2)
dw/dt = 1/[2(x^2+y^2+z^2)] * (2xx'+2yy'+2zz')
= ((9sin t)(9cos t)+(6cos t)(-6sin t)+(7tan t)(7sec^2 t))/(81sin^2 t + 36cos^2 t + 7tan^2 t)
= (45sint*cost + 49tant*sec^2 t)/(81sin^2 t + 36cos^2 t + 49tan^2 t)

Don't think you can simplify that much.
Similar Questions
  1. How do I use the chain rule to find the derivative ofsquare root(1-x^2) also, are there any general hints or tips for
    1. answers icon 1 answer
  2. Find the derivative of [(2+x)/(x-3)]^(2/5)I tried the power of a function rule, quotient rule, the chain rule but keep getting
    1. answers icon 2 answers
  3. Find the derivative of the function.y= xcosx - sinx What's the derivative? I get only one trig function. I will be happy to
    1. answers icon 0 answers
  4. How do you find f'(x), f''(x) of f(x) =ln (1-x)IF f(x) = ln (1-x) Then use the chain rule with u(x) = 1-x f(u) = ln u df/dx =
    1. answers icon 0 answers
more similar questions