Find the derivative of e 6 + x cos ⁡ ( π x 3 ) .

Question

Find the derivative of             e              6        +        x                    cos      ⁡      (                        π          x                3            )      .

Raelynn Parker · Accepted Answer

Differentiate using the Quotient Rule which states that&amp;nbsp;ddx[f(x)g(x)]&amp;nbsp;is&amp;nbsp;g(x)ddx[f(x)]-f(x)ddx[g(x)]g(x)2&amp;nbsp;where&amp;nbsp;f(x)=e6+x&amp;nbsp;and&amp;nbsp;g(x)=cos(πx3).cos(πx3)ddx[e6+x]-e6+xddx[cos(πx3)]cos2(πx3)Differentiate using the chain rule, which states that&amp;nbsp;ddx[f(g(x))]&amp;nbsp;is&amp;nbsp;f′(g(x))g′(x)&amp;nbsp;where&amp;nbsp;f(x)=ex&amp;nbsp;and&amp;nbsp;g(x)=6+x.cos(πx3)(e6+xddx[6+x])-e6+xddx[cos(πx3)]cos2(πx3)Differentiate.cos(πx3)e6+x-e6+xddx[cos(πx3)]cos2(πx3)Differentiate using the chain rule, which states that&amp;nbsp;ddx[f(g(x))]&amp;nbsp;is&amp;nbsp;f′(g(x))g′(x)&amp;nbsp;where&amp;nbsp;f(x)=cos(x)&amp;nbsp;and&amp;nbsp;g(x)=πx3.cos(πx3)e6+x-e6+x(-sin(πx3)ddx[πx3])cos2(πx3)Differentiate.cos(πx3)e6+x+πsin(πx3)e6+x3cos2(πx3)To write&amp;nbsp;cos(πx3)e6+x&amp;nbsp;as a fraction with a common denominator, multiply by&amp;nbsp;33.cos(πx3)e6+x⋅33+πsin(πx3)e6+x3cos2(πx3)Combine&amp;nbsp;cos(πx3)e6+x&amp;nbsp;and&amp;nbsp;33.cos(πx3)e6+x⋅33+πsin(πx3)e6+x3cos2(πx3)Combine the numerators over the common denominator.cos(πx3)e6+x⋅3+πsin(πx3)e6+x3cos2(πx3)Move&amp;nbsp;3&amp;nbsp;to the left of&amp;nbsp;cos(πx3)e6+x.3⋅(cos(πx3)e6+x)+πsin(πx3)e6+x3cos2(πx3)Rewrite&amp;nbsp;3cos(πx3)e6+x+πsin(πx3)e6+x3cos2(πx3)&amp;nbsp;as a product.3cos(πx3)e6+x+πsin(πx3)e6+x3⋅1cos2(πx3)Convert from&amp;nbsp;1cos2(πx3)&amp;nbsp;to&amp;nbsp;sec2(πx3).3cos(πx3)e6+x+πsin(πx3)e6+x3sec2(πx3)Combine&amp;nbsp;3cos(πx3)e6+x+πsin(πx3)e6+x3&amp;nbsp;and&amp;nbsp;sec2(πx3).(3cos(πx3)e6+x+πsin(πx3)e6+x)sec2(πx3)3Simplify.e6+x(3cos(πx3)+π&amp;nbsp;sin(πx3))3cos2(πx3)

Find the derivative of e 6 + x </mrow>

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