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How do you find the derivative of #f(x) = (cos x) e^(-xsqrt3)#?

Answer 1

Cameron Alexander

We can use the product rule, which states that, be #y=f(x)g(x)#, then

#(dy)/(dx)=f'(x)g(x)+f(x)g'(x)#

So, we need the elements:

#f(x)=cos(x)#

#f'(x)=-sen(x)#

#g(x)=e^(-x*sqrt(3))#

#g'(x)# demands chain rule. The chain rule states that the derivative is given by #(dy)/(dx)=(dy)/(du)(du)/(dx)#. Thus, renaming #u=-xsqrt(3)#, we have that:

#g'(x)=-sqrt(3)e^(-xsqrt(3))#

Solving the original derivation, then:

#(dy)/(dx)=-sen(x)e^(-xsqrt(3))+cos(x)(-sqrt(3)e^(-xsqrt(3)))#

As stated by a law of exponentials, #a^-n=1/a^n#

#(dy)/(dx)=-(sen(x))/e^(xsqrt(3))-(sqrt(3)cos(x))/e^(xsqrt(3))#

#(dy)/(dx)=-(sen(x)+sqrt(3)cos(x))/e^(xsqrt(3))#

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Answer 2

Christopher Allers

To find the derivative of ( f(x) = (\cos x) e^{-x\sqrt{3}} ), you can use the product rule along with the chain rule. The derivative is:

[ f'(x) = -\sqrt{3}(\cos x)e^{-x\sqrt{3}} + (\sin x)e^{-x\sqrt{3}} ]

This can be simplified as:

[ f'(x) = e^{-x\sqrt{3}} (\sin x - \sqrt{3} \cos x) ]

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Answer from HIX Tutor

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