What is the derivative of
[(3t+4)/(6t-7)]^3?
To find the derivative of the given function, we can use the chain rule and the power rule of differentiation.
First, let's rewrite the function as a power function:
f(t) = [(3t+4)/(6t-7)]^3
To apply the chain rule, we consider the derivative of the outer function, which is taking the power of 3. We multiply this by the derivative of the inner function.
Step 1: Find the derivative of the outer function.
Using the power rule, if we have f(t) = g(t)^n, where g(t) is a function of t and n is a constant, the derivative is given by:
f'(t) = n * g(t)^(n-1) * g'(t)
In this case, g(t) = (3t+4)/(6t-7) and n = 3.
So, f'(t) = 3 * [(3t+4)/(6t-7)]^2 * g'(t)
Step 2: Find the derivative of the inner function.
The inner function, g(t) = (3t+4)/(6t-7), is a division of two functions. We can differentiate each part separately.
Let's differentiate (3t+4) and (6t-7) separately.
Differentiating (3t+4):
The derivative of 3t+4 with respect to t is simply 3, as the derivative of a constant term (4) is zero.
Differentiating (6t-7):
The derivative of 6t-7 with respect to t is 6, as the derivative of a constant term (-7) is zero.
Step 3: Multiply the derivatives of the inner function with the derivative of the outer function.
We can now substitute the derivatives of the inner function and the outer function that we found in step 2 into the formula from step 1.
f'(t) = 3 * [(3t+4)/(6t-7)]^2 * g'(t)
= 3 * [(3t+4)/(6t-7)]^2 * [(3)(6) - (4)(6)]
= 3 * [(3t+4)/(6t-7)]^2 * (18 - 24)
= -18 * [(3t+4)/(6t-7)]^2
Therefore, the derivative of [(3t+4)/(6t-7)]^3 is -18 * [(3t+4)/(6t-7)]^2.