f(x)=7cosx+6tanx, find f'(x)and f'(5pie/4)
f'(x) = -7sin x = 6 sec^2 x
Plug in 5 pi/4 for x and see what you get for f'(5 pi/4)
the answer was 6sec^2x-7sinx but what would it be if you put in (5pi/4) how do u put sec^2 in the caculator
To find the derivative of a function, we can use the rules of differentiation. In this case, we'll use the sum, product, and chain rules.
Given the function f(x) = 7cos(x) + 6tan(x), we can find its derivative f'(x) as follows:
Step 1: Apply the sum rule.
The sum rule states that the derivative of the sum of two functions is equal to the sum of their individual derivatives.
Thus, the derivative of 7cos(x) + 6tan(x) is equal to the derivative of 7cos(x) plus the derivative of 6tan(x).
Step 2: Apply the product rule to each term.
The product rule states that the derivative of the product of two functions is equal to the first function multiplied by the derivative of the second function, plus the second function multiplied by the derivative of the first function.
Let's start with the first term, 7cos(x):
The derivative of cos(x) is -sin(x) (by the derivative of cosine formula).
Applying the product rule, we get: 7 * (-sin(x)) = -7sin(x).
Now, let's move on to the second term, 6tan(x):
The derivative of tan(x) is sec^2(x) (by the derivative of tangent formula).
Applying the product rule, we get: 6 * sec^2(x).
Thus, the derivative of f(x) = 7cos(x) + 6tan(x) is f'(x) = -7sin(x) + 6sec^2(x).
To find f'(5π/4), substitute x = 5π/4 into the derivative equation f'(x) = -7sin(x) + 6sec^2(x):
f'(5π/4) = -7sin(5π/4) + 6sec^2(5π/4).
Now, we can calculate the values using the trigonometric values of sin and sec at 5π/4:
sin(5π/4) = sin(π/4) = 1/√2 (since sin(π/4) = 1/√2)
sec^2(5π/4) = sec^2(π/4) = 2 (since sec^2(π/4) = 2)
Substituting these values, we get:
f'(5π/4) = -7 * (1/√2) + 6 * 2 = -7/√2 + 12.
Therefore, f'(5π/4) = -7/√2 + 12.