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Math  /  Calculus

Question2. Differentiate the following function. (Do not simplify your a) [8pts] f(x)=cos⁡(2x)tan⁡(3x)−ex2−3xf(x)=\cos (2 x) \tan (3 x)-e^{x^{2}-3 x}

Studdy Solution

STEP 1

What is this asking? Find the *derivative* of a function that's a *difference* between a *product* of trigonometric functions and an *exponential* function with a little polynomial inside. Watch out! Don't forget the *chain rule* and *product rule*!
Also, remember the derivatives of tan⁡(x)\tan(x), cos⁡(x)\cos(x), and exe^x.

STEP 2

1. Differentiate the product.
2. Differentiate the exponential.
3. Combine the results.

STEP 3

Alright, let's **tackle** this product first: cos⁡(2x)⋅tan⁡(3x)\cos(2x) \cdot \tan(3x).
We'll use the **product rule**, which says (f⋅g)′=f′⋅g+f⋅g′(f \cdot g)' = f' \cdot g + f \cdot g'.

STEP 4

Let f(x)=cos⁡(2x)f(x) = \cos(2x).
Then f′(x)=−sin⁡(2x)⋅2=−2sin⁡(2x)f'(x) = -\sin(2x) \cdot 2 = -2\sin(2x) by the **chain rule**.
Remember, the derivative of cos⁡(x)\cos(x) is −sin⁡(x)-\sin(x)!

STEP 5

Now, let g(x)=tan⁡(3x)g(x) = \tan(3x).
Then g′(x)=sec⁡2(3x)⋅3=3sec⁡2(3x)g'(x) = \sec^2(3x) \cdot 3 = 3\sec^2(3x) using the **chain rule** again.
The derivative of tan⁡(x)\tan(x) is sec⁡2(x)\sec^2(x)!

STEP 6

Putting it all together for the product rule: ddx(cos⁡(2x)tan⁡(3x))=−2sin⁡(2x)⋅tan⁡(3x)+cos⁡(2x)⋅3sec⁡2(3x)\frac{d}{dx}(\cos(2x)\tan(3x)) = -2\sin(2x) \cdot \tan(3x) + \cos(2x) \cdot 3\sec^2(3x) =−2sin⁡(2x)tan⁡(3x)+3cos⁡(2x)sec⁡2(3x).= -2\sin(2x)\tan(3x) + 3\cos(2x)\sec^2(3x).So, the derivative of the **first part** of our function is done!

STEP 7

Now, let's **conquer** the exponential part: ex2−3xe^{x^2 - 3x}.
We'll use the **chain rule** here.

STEP 8

The derivative of eue^u is eu⋅dudxe^u \cdot \frac{du}{dx}.
In our case, u=x2−3xu = x^2 - 3x, so dudx=2x−3\frac{du}{dx} = 2x - 3.

STEP 9

Therefore, the derivative of ex2−3xe^{x^2 - 3x} is ex2−3x⋅(2x−3)=(2x−3)ex2−3xe^{x^2 - 3x} \cdot (2x - 3) = (2x - 3)e^{x^2 - 3x}.
Awesome!

STEP 10

Remember, our original function was f(x)=cos⁡(2x)tan⁡(3x)−ex2−3xf(x) = \cos(2x)\tan(3x) - e^{x^2 - 3x}.
We've found the derivatives of both parts.
Now we just **subtract** them!

STEP 11

f′(x)=−2sin⁡(2x)tan⁡(3x)+3cos⁡(2x)sec⁡2(3x)−(2x−3)ex2−3x.f'(x) = -2\sin(2x)\tan(3x) + 3\cos(2x)\sec^2(3x) - (2x - 3)e^{x^2 - 3x}.

STEP 12

The derivative of f(x)f(x) is f′(x)=−2sin⁡(2x)tan⁡(3x)+3cos⁡(2x)sec⁡2(3x)−(2x−3)ex2−3xf'(x) = -2\sin(2x)\tan(3x) + 3\cos(2x)\sec^2(3x) - (2x - 3)e^{x^2 - 3x}.

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