(1 point) Suppose that f(x) = Te’ exe Find f ‘ (3). f'(3)
The Correct Answer and Explanation is:
To solve for f′(3)f'(3)f′(3), let’s go through the problem step by step.
We are given:
f(x)=7ex+xexf(x) = 7e^x + xe^xf(x)=7ex+xex
We need to find the derivative f′(x)f'(x)f′(x), then evaluate it at x=3x = 3x=3.
Step 1: Find the derivative f′(x)f'(x)f′(x)
The function f(x)f(x)f(x) has two terms: 7ex7e^x7ex and xexxe^xxex.
We will apply basic differentiation rules:
- The derivative of exe^xex is exe^xex
- The derivative of a constant multiplied by a function follows the constant rule
- For xexxe^xxex, we use the product rule:
ddx(uv)=u′v+uv′\frac{d}{dx} (uv) = u’v + uv’dxd(uv)=u′v+uv′
Differentiate each term:
- ddx(7ex)=7ex\frac{d}{dx} (7e^x) = 7e^xdxd(7ex)=7ex
- ddx(xex)\frac{d}{dx} (xe^x)dxd(xex) using product rule:
- u=xu = xu=x, so u′=1u’ = 1u′=1
- v=exv = e^xv=ex, so v′=exv’ = e^xv′=ex
Thus:
ddx(xex)=(1)(ex)+(x)(ex)=ex+xex\frac{d}{dx} (xe^x) = (1)(e^x) + (x)(e^x) = e^x + xe^xdxd(xex)=(1)(ex)+(x)(ex)=ex+xex
So, the full derivative is:
f′(x)=7ex+ex+xexf'(x) = 7e^x + e^x + xe^xf′(x)=7ex+ex+xex
We can factor terms:
f′(x)=7ex+ex+xex=(7ex+ex)+xex=8ex+xexf'(x) = 7e^x + e^x + xe^x = (7e^x + e^x) + xe^x = 8e^x + xe^xf′(x)=7ex+ex+xex=(7ex+ex)+xex=8ex+xex
Step 2: Evaluate f′(3)f'(3)f′(3)
We substitute x=3x = 3x=3 into the derivative:
f′(3)=8e3+3e3=(8+3)e3=11e3f'(3) = 8e^3 + 3e^3 = (8 + 3)e^3 = 11e^3f′(3)=8e3+3e3=(8+3)e3=11e3
Final Answer:
f′(3)=11e3f'(3) = 11e^3f′(3)=11e3
Explanation:
In this problem, we apply fundamental differentiation rules: constant multiples, the derivative of the exponential function, and the product rule. The first term 7ex7e^x7ex differentiates directly to 7ex7e^x7ex, while the second term xexxe^xxex requires the product rule, yielding ex+xexe^x + xe^xex+xex. Combining terms gives 8ex+xex8e^x + xe^x8ex+xex. Plugging in x=3x = 3x=3 provides the final result of 11e311e^311e3, representing the slope of the tangent line to the function f(x)f(x)f(x) at the point where x=3x = 3x=3. This derivative indicates how rapidly the function changes at that specific point.
