How do you find the derivative of Inverse trig function #y = (sin(3x) + cot(x^3))^8#?

Question

Luke Alvarez · Accepted Answer

#y^' = 24 * [sin(3x) + cot(x^3)]^7 * [cos(3x) - x^2csc^2(x^3)]#

From the looks of it, you're going to have to use the chain rule three times to differentiate this function.

Before you get started, keep in mind that

#d/dx(sinx) = cosx" "# and #" "d/dx(cotx) = -csc^2x#

So, use the chain rule first for #u^8#, with #u = sin(3x) + cot(x^3)#

#d/dx(y) = d/(du)(u^8) * d/dx(u)#

#y^' = 8u^7 * d/dx(sin(3x) + cot(x^3))#

You can differentiate this function, #d/dx(sin(3x) + cot(x^3))# by using the chain rule twice, once for #sint#, with #t = 3x#, and once more for #cotv#, with #v = x^3#.

This means that you can write

#d/dx(sint) = d/(dt)sint * d/dx(t)#

#d/dx(sint) = cost * d/dx(3x)#

#d/dx(sin(3x)) = cos(3x) * 3#

and

#d/dx(cotv) = d/(dv)cotv * d/dx(v)#

#d/dx(cotv) = -csc^2v * d/dx(x^3)#

#d/dx(cot(x^3)) = -csc^2(x^3) * 3x^2#

Plug these derivatives into the calculation for #y^'# to get

#y^' = 8u^7 * [3cos(3x) - 3x^2csc^2(x^3)]#

#y^' = 8[sin(3x) + cot(x^3)]^7 * 3 * [cos(3x) - x^2csc^2(x^3)]#

.#y^' = color(green)(24 * [sin(3x) + cot(x^3)]^7 * [cos(3x) - x^2csc^2(x^3)])#

Evelyn Agard · Answer

undefined To find the derivative of the given function $y = (\sin(3x) + \cot(x^3))^8$, we will use the chain rule, the product rule, and the derivatives of basic trigonometric functions. The chain rule is used when differentiating a composite function, which in this case is a function raised to a power. The product rule may be needed for differentiating expressions like $\cot(x)$, which is equivalent to $\frac{\cos(x)}{\sin(x)}$.

Given:
$$y = (\sin(3x) + \cot(x^3))^8$$

Let $u = \sin(3x) + \cot(x^3)$, so that $y = u^8$.

First, differentiate $y$ with respect to $u$ (using the power rule):
$$\frac{dy}{du} = 8u^7$$

Next, differentiate $u$ with respect to $x$:
$$u = \sin(3x) + \cot(x^3)$$
$$du/dx = \frac{d}{dx}[\sin(3x)] + \frac{d}{dx}[\cot(x^3)]$$

The derivative of $\sin(3x)$ with respect to $x$, using the chain rule, is:
$$3\cos(3x)$$

The derivative of $\cot(x^3)$ with respect to $x$, recognizing that $\cot(x) = \frac{1}{\tan(x)}$ or $\cos(x)/\sin(x)$, and applying the chain rule, is:
$$-\csc^2(x^3) \cdot 3x^2$$
Which simplifies to:
$$-3x^2\csc^2(x^3)$$

Thus:
$$du/dx = 3\cos(3x) - 3x^2\csc^2(x^3)$$

Finally, apply the chain rule by multiplying $\frac{dy}{du}$ and $\frac{du}{dx}$ to find $\frac{dy}{dx}$:
$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 8(\sin(3x) + \cot(x^3))^7 \cdot (3\cos(3x) - 3x^2\csc^2(x^3))$$

Therefore, the derivative of the given function with respect to $x$ is:
$$\frac{dy}{dx} = 8(\sin(3x) + \cot(x^3))^7 \cdot (3\cos(3x) - 3x^2\csc^2(x^3))$$