Derivative Math Example 5

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Example 5

hard
Find the derivative of f(x)=x3sinโก(x)โˆ’exx2f(x) = x^3 \sin(x) - \frac{e^x}{x^2} and evaluate fโ€ฒ(ฯ€)f'(\pi).

Solution

  1. 1
    Apply the product rule to x3sinโก(x)x^3 \sin(x): ddx[x3sinโก(x)]=3x2sinโก(x)+x3cosโก(x)\frac{d}{dx}[x^3 \sin(x)] = 3x^2 \sin(x) + x^3 \cos(x).
  2. 2
    Apply the quotient rule to exx2\frac{e^x}{x^2}: ddx[exx2]=exโ‹…x2โˆ’exโ‹…2xx4=ex(xโˆ’2)x3\frac{d}{dx}\left[\frac{e^x}{x^2}\right] = \frac{e^x \cdot x^2 - e^x \cdot 2x}{x^4} = \frac{e^x(x-2)}{x^3}.
  3. 3
    Combine: fโ€ฒ(x)=3x2sinโก(x)+x3cosโก(x)โˆ’ex(xโˆ’2)x3f'(x) = 3x^2 \sin(x) + x^3 \cos(x) - \frac{e^x(x-2)}{x^3}.
  4. 4
    Evaluate at x=ฯ€x = \pi: sinโก(ฯ€)=0\sin(\pi)=0, cosโก(ฯ€)=โˆ’1\cos(\pi)=-1, so fโ€ฒ(ฯ€)=0+ฯ€3(โˆ’1)โˆ’eฯ€(ฯ€โˆ’2)ฯ€3=โˆ’ฯ€3โˆ’eฯ€(ฯ€โˆ’2)ฯ€3f'(\pi) = 0 + \pi^3(-1) - \frac{e^\pi(\pi-2)}{\pi^3} = -\pi^3 - \frac{e^\pi(\pi-2)}{\pi^3}.

Answer

fโ€ฒ(ฯ€)=โˆ’ฯ€3โˆ’eฯ€(ฯ€โˆ’2)ฯ€3f'(\pi) = -\pi^3 - \frac{e^\pi(\pi - 2)}{\pi^3}
This problem requires applying both the product rule and quotient rule, then substituting x=ฯ€x=\pi and using the exact values sinโก(ฯ€)=0\sin(\pi)=0 and cosโก(ฯ€)=โˆ’1\cos(\pi)=-1 to obtain the final expression.

About Derivative

The instantaneous rate of change of a function at a single point, defined as the limit of the slope of secant lines.

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