Find the derivative of the function. h(t) = (t^3 /t^6 +3)^2?

The answer is as follows:

As written:

h(t) = (t³/t⁶ + 3)²

The first term reduces:

h(t) = (1/t³ + 3)²

Now we have the chain rule.  Setting that up:

h = u² and u = 1/t³ + 3

Let's put that into negative exponent form so we can see what we need to do in the next step:

h = u² and u = t⁻³ + 3

Next, the derivative of both:

dh/du = 2u and du/dt = -3t⁻⁴

Back to fraction form:

dh/du = 2u and du/dt = -3/t⁴

Now the chain rule:

dh/dt = dh/du * du/dt
dh/dt = 2u * (-3/t⁴)
dh/dt = -6u/t⁴

Now put the expression in terms of t back in for u:

h'(t) = -6(1/t³ + 3) / t⁴

Now we need to simplify that.  Start with getting a common denominator, then add the fractions, then distribute the -6:

h'(t) = -6(1/t³ + 3t³/t³) / t⁴
h'(t) = -6[(1 + 3t³) / t³] / t⁴
h'(t) = [(-6 - 18t³) / t³] / t⁴

Dividision is the multipllication of the reciprocal:

h'(t) = [(-6 - 18t³) / t³] * 1/t⁴

Multiply:

h'(t) = (-6 - 18t³) / t⁷

or:

h'(t) = -6/t⁷ - 18t³/t⁷
h'(t) = -6/t⁷ - 18/t⁴

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Now presuming you really meant:

h(t) = [t³ / (t⁶ + 3)]²

Now we have a quotient rule in the middle of this chain rule.

Let's start setting things up:

h = u² and u = t³ / (t⁶ + 3)

To get the derivative of u we need the quotient rule.  So let's set that up, then snake back:

u = y/z and y = t³ and z = t⁶ + 3
u' = (y'z - yz') / z² and y' = 3t² and z' = 6t⁵

Now we can substitute what we know:

u' = (y'z - yz') / z²
u' = [3t²(t⁶ + 3) - t³ * 6t⁵] / (t⁶ + 3)²

Simplify that:

u' = (3t⁸ + 9t² - 6t⁸) / (t⁶ + 3)²
u' = (-3t⁸ + 9t²) / (t⁶ + 3)²

So now we go back to our chain rule and get the dirivatives of the pieces:

h = u² and u = t³ / (t⁶ + 3)
dh/du = 2u and du/dt = (-3t⁸ + 9t²) / (t⁶ + 3)²

Chain rule:

dh/dt = dh/du * du/dt
dh/dt = 2u(-3t⁸ + 9t²) / (t⁶ + 3)²

Substitute the expression for u:

h'(t) = 2[t³ / (t⁶ + 3)](-3t⁸ + 9t²) / (t⁶ + 3)²

Simplify:

h'(t) = [2t³ / (t⁶ + 3)](-3t⁸ + 9t²) / (t⁶ + 3)²
h'(t) = [2t³(-3t⁸ + 9t²) / (t⁶ + 3)] / (t⁶ + 3)²
h'(t) = [(-6t¹¹ + 18t⁵) / (t⁶ + 3)] / (t⁶ + 3)²

Change to the multipliation of the reciprocal:

h'(t) = [(-6t¹¹ + 18t⁵) / (t⁶ + 3)] * 1/(t⁶ + 3)²
h'(t) = (-6t¹¹ + 18t⁵) / (t⁶ + 3)³