Suppose that f(π4)=−4 and f′(π4)=7, and let g(x)=f(x)sinx and h(x)=cosxf(x). Answer the following questions.
1. Find g′(π/4).
2. Find h′(π/4).
if g(x) = f(x)sinx, then
g'(x) = f(x)cosx + f'(x)sinx
g'(π/4) = f(π/4)cos(π/4) + f'(π/4)sin(π/4)
= -4(√2/2) + 7(√2/2) , since sinπ/4 = cosπ/4
= 3√2/2
repeat the same steps for h'(π/4)
To find g'(π/4), we can use the product rule.
1. Find g'(π/4):
First, let's find the derivative of f(x)sin(x) with respect to x using the product rule.
g(x) = f(x)sin(x)
g'(x) = f'(x)sin(x) + f(x)cos(x)
Now, substitute π/4 into the equation:
g'(π/4) = f'(π/4)sin(π/4) + f(π/4)cos(π/4)
Given that f(π/4) = -4 and f'(π/4) = 7, we can substitute these values into the equation:
g'(π/4) = (7)(√2/2) + (-4)(√2/2)
g'(π/4) = 7√2/2 - 4√2/2
g'(π/4) = (7 - 4)√2/2
g'(π/4) = 3√2/2
Therefore, g'(π/4) = 3√2/2.
2. Find h'(π/4):
The derivative of cos(x)f(x) with respect to x can be found using the product rule.
h(x) = cos(x)f(x)
h'(x) = -sin(x)f(x) + cos(x)f'(x)
Now, substitute π/4 into the equation:
h'(π/4) = -sin(π/4)f(π/4) + cos(π/4)f'(π/4)
Given that f(π/4) = -4 and f'(π/4) = 7, we can substitute these values into the equation:
h'(π/4) = -(-4)(√2/2) + (√2/2)(7)
h'(π/4) = 4√2/2 + 7√2/2
h'(π/4) = (4 + 7)√2/2
h'(π/4) = 11√2/2
Therefore, h'(π/4) = 11√2/2.
To find the derivatives of g(x) and h(x), we need to use the product rule and chain rule. Let's differentiate them step by step:
1. To find g'(π/4):
We have g(x) = f(x)sin(x).
Using the product rule,
g'(x) = f'(x)sin(x) + f(x)cos(x).
Now, substitute x = π/4 into the equation:
g'(π/4) = f'(π/4)sin(π/4) + f(π/4)cos(π/4).
We know that f(π/4) = -4 and f'(π/4) = 7, so let's substitute these values:
g'(π/4) = 7sin(π/4) + (-4)cos(π/4).
Using the values of sin(π/4) = 1/√2 and cos(π/4) = 1/√2, simplify the equation:
g'(π/4) = 7(1/√2) + (-4)(1/√2) = 7/√2 - 4/√2 = (7-4)/√2 = 3/√2.
Therefore, g'(π/4) = 3/√2.
2. To find h'(π/4):
We have h(x) = cos(x)f(x).
Using the product rule,
h'(x) = -sin(x)f(x) + cos(x)f'(x).
Now, substitute x = π/4 into the equation:
h'(π/4) = -sin(π/4)f(π/4) + cos(π/4)f'(π/4).
Substituting f(π/4) = -4 and f'(π/4) = 7, we have:
h'(π/4) = -sin(π/4)(-4) + cos(π/4)(7).
Using the values of sin(π/4) = 1/√2 and cos(π/4) = 1/√2, simplify the equation:
h'(π/4) = (1/√2)(-4) + (1/√2)(7) = -4/√2 + 7/√2 = (7-4)/√2 = 3/√2.
Therefore, h'(π/4) = 3/√2.
So, the answers to the questions are:
1. g'(π/4) = 3/√2.
2. h'(π/4) = 3/√2.