Question

all parts have been answered. differentiate ( f(x) = (2 + x^{10})^{2/3} ). ( f(x) = \frac{6x(2 + x^{10})^{-1/3}}{3} ) next item

Explanation:

Step1: Identify the function type

The function \( F(x)=(2 + x^{9})^{\frac{2}{3}} \) is a composite function, so we use the chain rule. The chain rule states that if \( y = u^n \) and \( u = g(x) \), then \( y'=n\cdot u^{n - 1}\cdot u' \). Here, \( u = 2+x^{9} \) and \( n=\frac{2}{3} \).

Step2: Apply the power rule to the outer function

First, differentiate the outer function with respect to \( u \). The derivative of \( u^n \) with respect to \( u \) is \( n\cdot u^{n - 1} \). So, for \( u = 2 + x^{9} \) and \( n=\frac{2}{3} \), we have:
\( \frac{d}{du}(u^{\frac{2}{3}})=\frac{2}{3}u^{\frac{2}{3}-1}=\frac{2}{3}u^{-\frac{1}{3}} \)

Step3: Differentiate the inner function with respect to \( x \)

Now, differentiate the inner function \( u = 2+x^{9} \) with respect to \( x \). The derivative of a constant (2) is 0, and the derivative of \( x^{9} \) with respect to \( x \) is \( 9x^{8} \) (by the power rule \( \frac{d}{dx}(x^k)=kx^{k - 1} \)). So, \( u'=\frac{d}{dx}(2 + x^{9})=9x^{8} \)

Step4: Multiply the results from Step2 and Step3 (Chain Rule)

By the chain rule, \( F'(x)=\frac{2}{3}u^{-\frac{1}{3}}\cdot9x^{8} \). Substitute back \( u = 2 + x^{9} \):
\( F'(x)=\frac{2}{3}(2 + x^{9})^{-\frac{1}{3}}\cdot9x^{8} \)
Simplify the coefficients: \( \frac{2}{3}\times9 = 6 \), so we get \( F'(x)=6x^{8}(2 + x^{9})^{-\frac{1}{3}} \)
We can rewrite \( (2 + x^{9})^{-\frac{1}{3}}=\frac{1}{(2 + x^{9})^{\frac{1}{3}}} \), but the form \( 6x^{8}(2 + x^{9})^{-\frac{1}{3}} \) (or equivalent simplified form) is the derivative. Wait, in the given answer, it's \( 6x(2 + x^{10})^{-\frac{1}{3}} \)? Wait, maybe there was a typo in the original function. If we assume the original function was \( F(x)=(2 + x^{10})^{\frac{2}{3}} \) (maybe a typo in the exponent, 10 instead of 9), then let's recalculate:

If \( F(x)=(2 + x^{10})^{\frac{2}{3}} \), then \( u = 2+x^{10} \), \( n=\frac{2}{3} \)

Step2: \( \frac{d}{du}(u^{\frac{2}{3}})=\frac{2}{3}u^{-\frac{1}{3}} \)

Step3: \( u'=\frac{d}{dx}(2 + x^{10}) = 10x^{9} \)? No, wait the given derivative has \( 6x \). Wait, maybe the original function is \( F(x)=(2 + x^{2})^{\frac{2}{3}} \)? No, the given derivative is \( 6x(2 + x^{10})^{-\frac{1}{3}} \). Wait, perhaps the original function is \( F(x)=(2 + x^{10})^{\frac{2}{3}} \), and when we apply the chain rule:

\( F'(x)=\frac{2}{3}(2 + x^{10})^{-\frac{1}{3}}\times10x^{9} \)? No, that doesn't match. Wait, the given derivative is \( 6x(2 + x^{10})^{-\frac{1}{3}} \), so let's see: if \( F(x)=(2 + x^{10})^{\frac{2}{3}} \), and maybe there was a miscalculation. Wait, maybe the original function is \( F(x)=(2 + x^{10})^{\frac{2}{3}} \), and the derivative steps:

Let \( u = 2 + x^{10} \), \( y = u^{\frac{2}{3}} \)

\( \frac{dy}{du}=\frac{2}{3}u^{-\frac{1}{3}} \)

\( \frac{du}{dx}=10x^{9} \)

Then \( \frac{dy}{dx}=\frac{2}{3}u^{-\frac{1}{3}}\times10x^{9}=\frac{20}{3}x^{9}(2 + x^{10})^{-\frac{1}{3}} \), which is not matching. Wait, the given derivative is \( 6x(2 + x^{10})^{-\frac{1}{3}} \), so maybe the original function is \( F(x)=(2 + x^{2})^{\frac{2}{3}} \)? No, \( x^2 \) derivative is \( 2x \), then \( \frac{2}{3}(2 + x^{2})^{-\frac{1}{3}}\times2x=\frac{4}{3}x(2 + x^{2})^{-\frac{1}{3}} \), not matching. Wait, maybe the original function is \( F(x)=(2 + x^{10})^{\frac{2}{3}} \) and there was a typo in the coefficient. Alternatively, maybe the original function is \( F(x)=(2 + x^{10})^{\frac{2}{3}} \) and the derivative is calculated as:

Wait, the given derivative is \( 6x(2 + x^{10})^{-\frac{1}{3}} \), let's reverse engineer. Let's suppose \( F'(x)=6x(2 + x^{10})^{-\frac{1}{3}} \), then integrate to find \( F(x) \).

Let \( u = 2 + x^{10} \), \( du = 10x^{9}dx \), no, not matching. Wait, if we consider \( F(x)=(2 + x^{10})^{\frac{2}{3}} \), then the derivative should be \( \frac{2}{3}(2 + x^{10})^{-\frac{1}{3}}\times10x^{9}=\frac{20}{3}x^{9}(2 + x^{10})^{-\frac{1}{3}} \). But the given derivative is \( 6x(2 + x^{10})^{-\frac{1}{3}} \), so maybe there was a typo in the original function's exponent. If the original function is \( F(x)=(2 + x^{2})^{\frac{2}{3}} \), derivative is \( \frac{2}{3}(2 + x^{2})^{-\frac{1}{3}}\times2x=\frac{4}{3}x(2 + x^{2})^{-\frac{1}{3}} \), still not. Alternatively, if the original function is \( F(x)=(2 + x^{10})^{\frac{2}{3}} \) and the coefficient is wrong, but according to the problem statement, we need to differentiate \( F(x)=(2 + x^{9})^{\frac{2}{3}} \) (assuming the original function's exponent is 9). Let's do that correctly:

\( F(x)=(2 + x^{9})^{\frac{2}{3}} \)

Let \( u = 2 + x^{9} \), \( y = u^{\frac{2}{3}} \)

\( \frac{dy}{du}=\frac{2}{3}u^{-\frac{1}{3}} \)

\( \frac{du}{dx}=9x^{8} \)

Then \( \frac{dy}{dx}=\frac{2}{3}u^{-\frac{1}{3}}\times9x^{8}=6x^{8}(2 + x^{9})^{-\frac{1}{3}} \)

Which can be written as \( \frac{6x^{8}}{(2 + x^{9})^{\frac{1}{3}}} \) or \( 6x^{8}(2 + x^{9})^{-\frac{1}{3}} \)

If we assume that there was a typo and the original function is \( F(x)=(2 + x^{10})^{\frac{2}{3}} \) and the derivative has a typo in the exponent of \( x \) (from \( x^{8} \) to \( x \) and \( x^{9} \) to \( x^{10} \)), but regardless, the correct differentiation using chain rule for \( F(x)=(2 + x^{n})^{\frac{2}{3}} \) is \( \frac{2}{3}(2 + x^{n})^{-\frac{1}{3}}\times n x^{n - 1} \)

In the given problem, if we take the original function as \( F(x)=(2 + x^{9})^{\frac{2}{3}} \), the derivative is \( 6x^{8}(2 + x^{9})^{-\frac{1}{3}} \) (which is equivalent to \( \frac{6x^{8}}{(2 + x^{9})^{\frac{1}{3}}} \) or \( 6x^{8}(2 + x^{9})^{-\frac{1}{3}} \))

Answer:

The derivative of \( F(x)=(2 + x^{9})^{\frac{2}{3}} \) is \( \boldsymbol{6x^{8}(2 + x^{9})^{-\frac{1}{3}}} \) (or equivalent forms like \( \frac{6x^{8}}{\sqrt[3]{2 + x^{9}}} \))