Find the second derivative of: h(t)= (t^2 + 5)^8
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
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
To find the second derivative of the function h(t) = (t^2 + 5)^8, we'll need to use the chain rule twice.
Step 1: Find the first derivative of h(t).
Let's start by finding the first derivative of h(t). Apply the chain rule to the function h(t) = (t^2 + 5)^8.
First, rewrite h(t) as u^8, where u = t^2 + 5.
So, h(t) = u^8.
Now, differentiate h(t) with respect to u, treating u as the independent variable.
dh/du = 8u^7.
To find the derivative of u with respect to t, take the derivative of u = t^2 + 5:
du/dt = 2t.
Now, using the chain rule, we can find the first derivative of h(t) with respect to t:
dh/dt = (dh/du) * (du/dt)
= 8u^7 * 2t
= 16tu^7.
Step 2: Find the second derivative of h(t).
To find the second derivative of h(t), we need to differentiate the first derivative we just found.
Take the derivative of dh/dt = 16tu^7 with respect to t:
d²h/dt² = d(16tu^7)/dt.
Since the derivative of 16tu^7 with respect to t is simply 16u^7, we have:
d²h/dt² = 16u^7.
Now substitute back in u = t^2 + 5:
d²h/dt² = 16(t^2 + 5)^7.
Therefore, the second derivative of h(t) = (t^2 + 5)^8 is d²h/dt² = 16(t^2 + 5)^7.