Find the derivative of the function. $$ g(u)=\left(\frac{u^{3}-3}{u^{3}+3}\right)^{7} $$ $$ g^{\prime}(u)=\square $$ Need Help? Read it

Question

UpStudy AI · Accepted Answer

Let
$$
g(u)=\left(\frac{u^3-3}{u^3+3}\right)^7.
$$
Define
$$
T(u)=\frac{u^3-3}{u^3+3},
$$
so that
$$
g(u) = \left(T(u)\right)^7.
$$

**Step 1. Differentiate using the chain rule**

Using the chain rule, we have
$$
g'(u)=7\left(T(u)\right)^6\cdot T'(u).
$$

**Step 2. Differentiate $$T(u)$$ using the quotient rule**

The quotient rule gives
$$
T'(u)=\frac{(u^3+3)\cdot (u^3-3)' - (u^3-3)\cdot (u^3+3)'}{(u^3+3)^2}.
$$

First, compute the derivatives:
$$
(u^3-3)'=3u^2,\quad (u^3+3)'=3u^2.
$$

Substitute these derivatives into the quotient rule:
$$
T'(u)=\frac{(u^3+3)\cdot 3u^2 - (u^3-3)\cdot 3u^2}{(u^3+3)^2}.
$$

Factor out $$3u^2$$:
$$
T'(u)=\frac{3u^2\left[(u^3+3) - (u^3-3)\right]}{(u^3+3)^2}.
$$

Simplify the numerator:
$$
(u^3+3)-(u^3-3)=6,
$$
so
$$
T'(u)=\frac{3u^2\cdot 6}{(u^3+3)^2}=\frac{18u^2}{(u^3+3)^2}.
$$

**Step 3. Combine the results**

Substitute $$T'(u)$$ into the expression for $$g'(u)$$:
$$
g'(u)=7\left(\frac{u^3-3}{u^3+3}\right)^6\cdot \frac{18u^2}{(u^3+3)^2}.
$$

Multiply the constants:
$$
7\cdot 18=126,
$$
so
$$
g'(u)=\frac{126u^2\left(u^3-3\right)^6}{\left(u^3+3\right)^8}.
$$

Thus, the derivative is
$$
g'(u)=\frac{126u^2\left(u^3-3\right)^6}{\left(u^3+3\right)^8}.
$$
The derivative of the function $$ g(u) = \left(\frac{u^3 - 3}{u^3 + 3}\right)^7 $$ is: $$ g'(u) = \frac{126u^2\left(u^3 - 3\right)^6}{\left(u^3 + 3\right)^8} $$

Find the derivative of the function.

Need Help? Read it

Upstudy AI Solution

Answer

Solution

Mind Expander

Related Questions

Latest Calculus Questions

Find the derivative of the function. Need Help? Read it