Suppose f(pi/3) = 3 and f '(pi/3) = −7,
and let g(x) = f(x) sin x and
h(x) = (cos x)/f(x). Find the following.
a. g'(pi/3)
b. h'(pi/3)
g = f*sin(x)
g' = f'*sin(x) + f*cos(x)
g'(pi/3) = (-7)*√3/2 + 3*1/2 = (3-7√3)/2
h = cos(x)/f
h' = (-sin(x)*f - cos(x)*f')/f^2
h'(pi/3) = [(-√3/2)*3 - (1/2)(-7)]/9 = (7-3√3)/18
Thank you very much sir and Good Day!
a. To find g'(pi/3), we need to apply the product rule. Recall that the product rule states that if we have two functions u(x) and v(x), then the derivative of their product w(x) = u(x) * v(x) is given by w'(x) = u'(x) * v(x) + u(x) * v'(x).
In this case, u(x) = f(x) and v(x) = sin(x). Therefore, u'(x) = f'(x) and v'(x) = cos(x).
Now, let's substitute the given values: u(pi/3) = f(pi/3) = 3, and v(pi/3) = sin(pi/3) = (√3)/2.
Applying the product rule, we have:
g'(x) = u'(x) * v(x) + u(x) * v'(x)
g'(pi/3) = u'(pi/3) * v(pi/3) + u(pi/3) * v'(pi/3)
g'(pi/3) = f'(pi/3) * sin(pi/3) + f(pi/3) * cos(pi/3)
Substituting the given values, we get:
g'(pi/3) = (-7) * (√3)/2 + 3 * cos(pi/3)
Now, simplify:
g'(pi/3) = (-7√3)/2 + 3 * 1/2
g'(pi/3) = -(7√3)/2 + 3/2
g'(pi/3) = (3-7√3)/2
Therefore, g'(pi/3) is equal to (3-7√3)/2.
b. To find h'(pi/3), we can again use the quotient rule. The quotient rule states that if we have two functions u(x) and v(x), then the derivative of the quotient w(x) = u(x) / v(x) is given by w'(x) = (u'(x) * v(x) - u(x) * v'(x)) / v^2(x).
In this case, u(x) = cos(x) and v(x) = f(x). Therefore, u'(x) = -sin(x) and v'(x) = f'(x).
Now, let's substitute the given values: u(pi/3) = cos(pi/3) = 1/2, and v(pi/3) = f(pi/3) = 3.
Applying the quotient rule, we have:
h'(x) = (u'(x) * v(x) - u(x) * v'(x)) / v^2(x)
h'(pi/3) = (u'(pi/3) * v(pi/3) - u(pi/3) * v'(pi/3)) / (v(pi/3))^2
h'(pi/3) = (-sin(pi/3) * 3 - (1/2) * (-7)) / (3)^2
Substituting the given values, we get:
h'(pi/3) = -(√3)/2 + 7/6
h'(pi/3) = (7 - 3√3)/6
Therefore, h'(pi/3) is equal to (7 - 3√3)/6.
To find the derivatives of g(x) and h(x), we can use the product rule and quotient rule respectively. Let's solve each part step-by-step.
a. To find g'(pi/3):
Step 1: Apply the product rule to differentiate g(x) = f(x) sin(x).
g'(x) = f'(x) sin(x) + f(x) cos(x)
Step 2: Substitute x = pi/3.
g'(pi/3) = f'(pi/3) sin(pi/3) + f(pi/3) cos(pi/3)
Given information:
f(pi/3) = 3
f'(pi/3) = -7
Step 3: Substitute the values into the equation.
g'(pi/3) = (-7) sin(pi/3) + 3 cos(pi/3)
Step 4: Simplify the expression.
g'(pi/3) = -7(sqrt(3)/2) + 3(1/2)
= -7(sqrt(3)/2) + 3/2
So, g'(pi/3) = -7(sqrt(3)/2) + 3/2.
b. To find h'(pi/3):
Step 1: Apply the quotient rule to differentiate h(x) = (cos(x)) / f(x).
h'(x) = (f(x)(-sin(x)) - cos(x)f'(x)) / (f(x))^2
Step 2: Substitute x = pi/3.
h'(pi/3) = (f(pi/3)(-sin(pi/3)) - cos(pi/3)f'(pi/3)) / (f(pi/3))^2
Given information:
f(pi/3) = 3
f'(pi/3) = -7
Step 3: Substitute the values into the equation.
h'(pi/3) = (3(-sin(pi/3)) - cos(pi/3)(-7)) / (3^2)
Step 4: Simplify the expression.
h'(pi/3) = (-3(sqrt(3))/2 + 7cos(pi/3)) / 9
= (-3(sqrt(3))/2 + 7(1/2)) / 9
= (-3(sqrt(3))/2 + 7/2) / 9
So, h'(pi/3) = (-3(sqrt(3))/2 + 7/2) / 9.
To find the values of g'(pi/3) and h'(pi/3), we will use the product rule and quotient rule, respectively.
a. To find g'(pi/3), we need to use the product rule. The product rule states that if we have two functions u(x) and v(x), then the derivative of their product is given by:
(uv)'(x) = u'(x)v(x) + u(x)v'(x)
In this case, u(x) = f(x) and v(x) = sin(x), so we have:
g(x) = f(x) * sin(x)
To find the derivative of g(x), we can apply the product rule:
g'(x) = f'(x) * sin(x) + f(x) * cos(x)
Now, to find g'(pi/3), we substitute x = pi/3 into the equation:
g'(pi/3) = f'(pi/3) * sin(pi/3) + f(pi/3) * cos(pi/3)
Substituting the given values:
f(pi/3) = 3 and f'(pi/3) = -7, we get:
g'(pi/3) = -7 * sin(pi/3) + 3 * cos(pi/3)
Calculating the values of sin(pi/3) and cos(pi/3), which are well-known, we have:
sin(pi/3) = sqrt(3)/2 and cos(pi/3) = 1/2.
So, substituting these values into the equation:
g'(pi/3) = -7 * (sqrt(3)/2) + 3 * (1/2)
Simplifying the expression:
g'(pi/3) = -7sqrt(3)/2 + 3/2
Thus, g'(pi/3) = (-7sqrt(3) + 3)/2.
b. To find h'(pi/3), we need to use the quotient rule. The quotient rule states that if we have two functions u(x) and v(x), then the derivative of their quotient is given by:
(h(x))' = (u'(x)v(x) - u(x)v'(x))/(v(x))^2
In this case, u(x) = cos(x) and v(x) = f(x), so we have:
h(x) = (cos(x))/f(x)
Applying the quotient rule, we get:
h'(x) = (f(x)(-sin(x)) - cos(x)f'(x))/(f(x))^2
Now, to find h'(pi/3), we substitute x = pi/3 into the equation:
h'(pi/3) = (f(pi/3)(-sin(pi/3)) - cos(pi/3)f'(pi/3))/(f(pi/3))^2
And substituting the given values:
f(pi/3) = 3 and f'(pi/3) = -7, we have:
h'(pi/3) = (3(-sin(pi/3)) - cos(pi/3)(-7))/3^2
Again using the values of sin(pi/3) and cos(pi/3):
sin(pi/3) = sqrt(3)/2 and cos(pi/3) = 1/2.
Substituting these values into the equation:
h'(pi/3) = 3(-sqrt(3)/2) - (1/2)(-7))/(3^2)
Simplifying the expression:
h'(pi/3) = (-3sqrt(3) + 7)/9.
Therefore, h'(pi/3) = (-3sqrt(3) + 7)/9.