Determine the slope of the tangent to the function f(x)=5e^x - 2e^(2x) at the point with x-coordinate x=1
To determine the slope of the tangent to the function f(x) at the point with x-coordinate x=1, we need to find the derivative of the function and evaluate it at x=1.
Step 1: Find the derivative of f(x)
To find the derivative of f(x), we can use the rules of differentiation. Since f(x) is a combination of exponential functions, we can use the chain rule.
The derivative of 5e^x is 5e^x
The derivative of -2e^(2x) is -4e^(2x) (apply the chain rule: derivative of e^(2x) is 2e^(2x) multiplied by the derivative of 2x, which is 2)
So, the derivative of f(x) is f'(x) = 5e^x - 4e^(2x)
Step 2: Evaluate the derivative at x=1
To evaluate the derivative at x=1, we substitute x=1 into the derivative expression.
f'(1) = 5e^1 - 4e^(2*1)
= 5e - 4e^2
The slope of the tangent to the function f(x) at x=1 is 5e - 4e^2.
To determine the slope of the tangent to the function f(x) at a specific point, we need to find the derivative of the function and evaluate it at that point.
Step 1: Find the derivative of f(x):
To find the derivative of f(x), we use the rules of differentiation. The derivative of each term in f(x) can be found using the chain rule. The derivative of 5e^x is 5e^x, and the derivative of -2e^(2x) is -4e^(2x) since the derivative of e^(kx) is ke^(kx).
So, the derivative of f(x) is:
f'(x) = 5e^x - 4e^(2x)
Step 2: Evaluate the derivative at x=1:
To find the slope of the tangent at x=1, we substitute x=1 into the derivative f'(x).
f'(1) = 5e^1 - 4e^(2*1)
= 5e - 4e^2
Therefore, the slope of the tangent to the function f(x) at x=1 is 5e - 4e^2.
well, f' = 5e^x - 4e^(2x)
so, plug in x=1