Question

Find the Derivative. f(x) = -xcos3x

Answer 1

Answer:

$\frac{d}{dx}(-xcos3x)= 3x \ sin(3x)- cos(3x)$

Step-by-step explanation:

Use the Product Rule for derivatives, which states that:

• $\frac{d}{dx} =[f(x)g(x)]= f(x)g'(x)+f'(x)g(x)$

In the function we are given, $f(x)=-x \cos3x$, we can break it up into two factors: -x is being multiplied by cos3x.

Now, we have the factors:

• $-x \\ cos 3x$

Before using the product rule, let's find the derivative of cos3x using the chain rule and the power rule.

• $\frac{d}{dx}(cos3x)= \frac{d}{dx} (cos3x) \times \frac{d}{dx} (3x) \\ \frac{d}{dx}(cos3x)= (-sin3x) \times 3 \\ \frac{d}{dx} (cos3x) = -3sin(3x)$

Now let's apply the product rule to f(x) = -xcos3x.

• $\frac{d}{dx}(-xcos3x)= (-x)(-3sin3x) + (-1)(cos3x)$

Simplify this equation.

• $\frac{d}{dx}(-xcos3x)= (x)(3sin3x) - (cos3x)$

Multiply x and 3 together and remove the parentheses.

• $\frac{d}{dx}(-xcos3x)= 3x \ sin(3x)- cos(3x)$

Therefore, this is the derivative of the function $f(x)=-xcos3x$.

Answer 2

Hi there!

$\large\boxed{f'(x) = -cos3x + 3xsin3x}$

$f(x) = -xcox3x\\\\\text{Use the product rule and chain rule to solve for the derivative:}\\\\\frac{dy}{dx}= -cos3x + (-x * (-sin3x) * 3)\\\\= -cos3x + 3xsin3x$