What is the equation of the tangent line of f(x)=cosx-e^xsinx at x=pi/3?

1 Answer

Equation of the tangent line
y-1/2+ sqrt(3)/2*e^(pi/3)=-1/2(sqrt(3)+e^(pi/3)+sqrt(3)e^(pi/3))(x-pi/3)

Explanation:

We start from the given equation f(x)=cos x-e^x sin x

Let us solve for the point of tangency first

f(pi/3)=cos (pi/3)-e^(pi/3) sin (pi/3)

f(pi/3)=1/2-e^(pi/3) sqrt(3)/2

Let us solve for the slope m now

f(x)=cos x-e^x sin x

Find the first derivative first

f' (x)=d/dx(cos x-e^x sin x)

f' (x)=-sin x-[e^x*cos x+sin x*e^x*1]

Slope m=f' (pi/3)=-sin (pi/3)-[e^(pi/3) cos (pi/3)+sin (pi/3)*e^(pi/3)]

m=f' (pi/3)=-sqrt(3)/2-[e^(pi/3) *1/2+sqrt(3)/2*e^(pi/3)]

m=f' (pi/3)=-sqrt(3)/2-[1/2+sqrt(3)/2]*e^(pi/3)

m=f' (pi/3)=-1/2[sqrt(3)+e^(pi/3)+sqrt(3)e^(pi/3)]*

Our Tangent Line:

y-f(pi/3)=m(x-pi/3)

y-1/2+ sqrt(3)/2*e^(pi/3)=-1/2(sqrt(3)+e^(pi/3)+sqrt(3)e^(pi/3))(x-pi/3)

Kindly see the graph of f(x)= cos x-e^x sin x and the tangent line
y-1/2+ sqrt(3)/2*e^(pi/3)=-1/2(sqrt(3)+e^(pi/3)+sqrt(3)e^(pi/3))(x-pi/3)
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God bless....I hope the explanation is useful.