Find dy/dx y=(x^3-1)cos(x)

$y = (x^{3} - 1) \cos (x)$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (y) = \frac{d}{d x} ((x^{3} - 1) \cos (x))$

Step 2

The derivative of $y$ with respect to $x$ is $y'$ .

$y'$

Step 3

Differentiate the right side of the equation.

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x^{3} - 1$ and $g (x) = \cos (x)$ .

$(x^{3} - 1) \frac{d}{d x} [\cos (x)] + \cos (x) \frac{d}{d x} [x^{3} - 1]$

The derivative of $\cos (x)$ with respect to $x$ is $- \sin (x)$ .

$(x^{3} - 1) (- \sin (x)) + \cos (x) \frac{d}{d x} [x^{3} - 1]$

Differentiate.

By the Sum Rule, the derivative of $x^{3} - 1$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 1]$ .

$(x^{3} - 1) (- \sin (x)) + \cos (x) (\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 1])$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$(x^{3} - 1) (- \sin (x)) + \cos (x) (3 x^{2} + \frac{d}{d x} [- 1])$

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$(x^{3} - 1) (- \sin (x)) + \cos (x) (3 x^{2} + 0)$

Add $3 x^{2}$ and $0$ .

$(x^{3} - 1) (- \sin (x)) + \cos (x) (3 x^{2})$

Simplify.

Apply the distributive property.

$x^{3} (- \sin (x)) - 1 (- \sin (x)) + \cos (x) (3 x^{2})$

Combine terms.

Multiply $- 1$ by $- 1$ .

$x^{3} (- \sin (x)) + 1 \sin (x) + \cos (x) (3 x^{2})$

Multiply $\sin (x)$ by $1$ .

$x^{3} (- \sin (x)) + \sin (x) + \cos (x) (3 x^{2})$

Reorder terms.

$- x^{3} \sin (x) + 3 x^{2} \cos (x) + \sin (x)$

Step 4

Reform the equation by setting the left side equal to the right side.

$y' = - x^{3} \sin (x) + 3 x^{2} \cos (x) + \sin (x)$

Step 5

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = - x^{3} \sin (x) + 3 x^{2} \cos (x) + \sin (x)$

Find dy/dx y=(x^3-1)cos(x)

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