Use the chain rule, or substitute
y=(t^2+5), and
dy/dt = 2t in
h(t)= (t^2 + 5)^8
to give
h(y)=y^8
and
h'(t)=dh/dy * dy/dt
Find the second derivative of: h(t)= (t^2 + 5)^8
3 answers
I'm not following. For the first derivative I got 8(t^2 + 5)^7 x (2t). I get stuck after that.
What I did was for the first derivative.
You can proceed in a similar way for the second derivative, but I guess substitution is out.
Take the derivative of what you've got as a product.
u=(t^2 + 5)^7
v=16t
d(uv)/dt = udv/dt + vdu/dt
You can proceed in a similar way for the second derivative, but I guess substitution is out.
Take the derivative of what you've got as a product.
u=(t^2 + 5)^7
v=16t
d(uv)/dt = udv/dt + vdu/dt
1 answer