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Question
calculate the derivative of the function. ( r(x) = (0.1x^2 - 4.5x + 2.6)^{2.6} ) ( r(x) = ) attempt 1: 10 attempts remaining. submit answer next item
Step1: Identify the outer and inner functions
Let \( u = 0.1x^{2}-4.5x + 2.6 \) (inner function) and \( r(u)=u^{2.6} \) (outer function). We will use the chain rule, which states that if \( r(x)=f(g(x)) \), then \( r^{\prime}(x)=f^{\prime}(g(x))\cdot g^{\prime}(x) \).
Step2: Differentiate the outer function
Differentiate \( r(u) = u^{2.6} \) with respect to \( u \) using the power rule \( \frac{d}{du}(u^{n})=n\cdot u^{n - 1} \). So, \( r^{\prime}(u)=2.6u^{2.6-1}=2.6u^{1.6} \).
Step3: Differentiate the inner function
Differentiate \( u = 0.1x^{2}-4.5x + 2.6 \) with respect to \( x \). Using the power rule for each term:
- The derivative of \( 0.1x^{2} \) is \( 0.1\times2x = 0.2x \)
- The derivative of \( - 4.5x \) is \( - 4.5 \)
- The derivative of the constant \( 2.6 \) is \( 0 \)
So, \( u^{\prime}(x)=0.2x - 4.5 \)
Step4: Apply the chain rule
Substitute \( u = 0.1x^{2}-4.5x + 2.6 \) into \( r^{\prime}(u) \) and multiply by \( u^{\prime}(x) \):
\( r^{\prime}(x)=2.6(0.1x^{2}-4.5x + 2.6)^{1.6}\cdot(0.2x - 4.5) \)
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\( 2.6(0.1x^{2}-4.5x + 2.6)^{1.6}(0.2x - 4.5) \)