Question

differentiate f(x) = (8 + x⁴)^(2/3). f(x) =

Explanation:

Step1: Identify the function type

The function \( F(x) = (8 + x^4)^{\frac{2}{3}} \) is a composite function, so we use the chain rule. The chain rule states that if \( F(x) = g(h(x)) \), then \( F'(x) = g'(h(x)) \cdot h'(x) \). Let \( u = 8 + x^4 \) (so \( h(x)=u \)) and \( g(u)=u^{\frac{2}{3}} \).

Step2: Differentiate the outer function

Differentiate \( g(u) = u^{\frac{2}{3}} \) with respect to \( u \) using the power rule \( \frac{d}{du}(u^n)=nu^{n - 1} \). Here, \( n=\frac{2}{3} \), so \( g'(u)=\frac{2}{3}u^{\frac{2}{3}-1}=\frac{2}{3}u^{-\frac{1}{3}} \). Substitute back \( u = 8 + x^4 \), we get \( g'(h(x))=\frac{2}{3}(8 + x^4)^{-\frac{1}{3}} \).

Step3: Differentiate the inner function

Differentiate \( h(x)=8 + x^4 \) with respect to \( x \). The derivative of a constant (8) is 0, and the derivative of \( x^4 \) using the power rule is \( 4x^{3} \). So \( h'(x)=4x^{3} \).

Step4: Apply the chain rule

Multiply \( g'(h(x)) \) and \( h'(x) \) together: \( F'(x)=\frac{2}{3}(8 + x^4)^{-\frac{1}{3}}\cdot4x^{3} \). Simplify the expression: \( F'(x)=\frac{8x^{3}}{3(8 + x^4)^{\frac{1}{3}}} \) (since \( a^{-n}=\frac{1}{a^{n}} \)).

Answer:

\( \dfrac{8x^{3}}{3(8 + x^{4})^{\frac{1}{3}}} \)