the slope of the line tangent to the curve y=(tanx/cosx) at x=(pi/3) is?
a)-8
b)2√3
c)3√2
d)8
e)14
y'(x) = (sec^2*cos + tan*sin)/cos^2
y'(pi/3) = (4*(1/2) + √3 * √3/2)/(1/4) = 14
To find the slope of the line tangent to the curve at a specific point, we can use calculus. The slope of the tangent line is the derivative of the function at that point.
The given function is y = (tanx / cosx). To find the derivative, we need to use the quotient rule of differentiation.
The quotient rule states that if we have a function of the form f(x) = g(x) / h(x), then the derivative is given by:
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
First, let's find the derivatives of the numerator and denominator separately.
1. Derivative of tanx:
The derivative of tanx is sec^2(x).
2. Derivative of cosx:
The derivative of cosx is -sinx.
Now, let's put these derivatives into the quotient rule formula:
f'(x) = (sec^2(x) * cosx - tanx * -sinx) / (cosx)^2
Simplifying further:
f'(x) = (sec^2(x) * cosx + sinx * tanx) / (cosx)^2
Now, we can substitute x = pi/3 into the derivative equation to find the slope at x = pi/3:
f'(pi/3) = (sec^2(pi/3) * cos(pi/3) + sin(pi/3) * tan(pi/3)) / (cos(pi/3))^2
Now, let's calculate each term separately:
sec^2(pi/3) = 1/cos^2(pi/3) = 1 / (1/2)^2 = 1 / (1/4) = 4
cos(pi/3) = 1/2
sin(pi/3) = √3/2
tan(pi/3) = sin(pi/3) / cos(pi/3) = (√3/2) / (1/2) = √3/1 = √3
Now, substitute these values back into the equation:
f'(pi/3) = (4 * (1/2) + (√3/2) * √3) / ((1/2)^2)
= (2 + 3) / (1/4)
= 5 / (1/4)
= 20
So, the slope of the line tangent to the curve at x = pi/3 is 20.
None of the given answer choices match this result, so it seems there might be a mistake in the question or the answer choices provided.