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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

Ask by Bowen Lawson. in the United States
Mar 15,2025

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

Solution

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 \[ g'(u)=\frac{126u^2\left(u^3-3\right)^6}{\left(u^3+3\right)^8}. \]

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To find the derivative of the function \( g(u)=\left(\frac{u^{3}-3}{u^{3}+3}\right)^{7} \), we can use the chain rule. Let \( h(u) = \frac{u^{3}-3}{u^{3}+3} \), then \( g(u) = h(u)^7 \). The derivative will be: \[ g^{\prime}(u) = 7h(u)^6 \cdot h^{\prime}(u) \] Next, we need to compute \( h^{\prime}(u) \) using the quotient rule: \[ h^{\prime}(u) = \frac{(3u^2)(u^3 + 3) - (u^3 - 3)(3u^2)}{(u^3 + 3)^2} \] Simplifying gives: \[ h^{\prime}(u) = \frac{(3u^2)(u^3 + 3) - (3u^2)(u^3 - 3)}{(u^3 + 3)^2} = \frac{6u^2 \cdot 3}{(u^3 + 3)^2} = \frac{18u^2}{(u^3 + 3)^2} \] Putting it all together, we get: \[ g^{\prime}(u) = 7 \left(\frac{u^3 - 3}{u^3 + 3}\right)^6 \cdot \frac{18u^2}{(u^3 + 3)^2} \] So the final derivative is: \[ g^{\prime}(u) = \frac{126u^2 \left( \frac{u^3-3}{u^3+3} \right)^6}{(u^3 + 3)^2} \]

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