Question

calculate the derivative of the function.

$r(x)=(0.2x^{2}-4.9x + 2.6)^{2.6}$

$r(x)=$

Explanation:

Step1: Apply chain - rule

Let $u = 0.2x^{2}-4.9x + 2.6$, then $r(x)=u^{2.6}$. The chain - rule states that $r^{\prime}(x)=\frac{dr}{du}\cdot\frac{du}{dx}$.
First, find $\frac{dr}{du}$:
Using the power rule $\frac{d}{du}(u^{n})=nu^{n - 1}$, for $n = 2.6$, we have $\frac{dr}{du}=2.6u^{2.6-1}=2.6u^{1.6}$.

Step2: Find $\frac{du}{dx}$

Differentiate $u = 0.2x^{2}-4.9x + 2.6$ with respect to $x$. Using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we get $\frac{du}{dx}=0.2\times2x-4.9=0.4x - 4.9$.

Step3: Calculate $r^{\prime}(x)$

Substitute $u = 0.2x^{2}-4.9x + 2.6$ back into $\frac{dr}{du}$ and then use the chain - rule formula:
$r^{\prime}(x)=\frac{dr}{du}\cdot\frac{du}{dx}=2.6(0.2x^{2}-4.9x + 2.6)^{1.6}(0.4x - 4.9)$

Answer:

$2.6(0.4x - 4.9)(0.2x^{2}-4.9x + 2.6)^{1.6}$