Deriving y= [x+(x+(sin(x)2))5]3

[grollio]

Find the derivative of y = [x + (x + (sin(x)²))⁵]³


I know that power and chain rule combined uses the equation
n[g(x)]^n-1 * g'(x)

I don't even really know where to start with so many layers in the equation. I can only find examples with only one power. with my attempt I got

3(5x+(sin(x))²)⁶ * 2sin(x)cos(x)+1

 

[jgens]

Your resulting expression for the derivative doesn't look correct. These kinds of problems with derivatives are actually very simple if you have some patience and remain consistent with notation. I'll start you off. Let u_1 = [x + (x + sin^2(x))^5] then we have that y = u_1^3. Taking the derivative with respect to x we find that,

\frac{dy}{dx} = 3u_1 * \frac{du_1}{dx} = 3u_1 \left [\frac{d}{dx}x + \frac{d}{dx}(x + sin^2(x))^5 \right ]

Now let u_2 = x + sin^2(x) and try to evaluate the rest from here on out.

 

[Hurkyl]

I don't even really know where to start with so many layers in the equation.

Don't be intimindated -- just work one layer at a time.

As you work through the calculation, you may find it useful to give temporary names to subexpressions (as jgens has done) to help you focus on the part you're working on.

 

[Дьявол]

Find the derivative of y = [x + (x + (sin(x)²))⁵]³


I know that power and chain rule combined uses the equation
n[g(x)]^n-1 * g'(x)

I don't even really know where to start with so many layers in the equation. I can only find examples with only one power. with my attempt I got

3(5x+(sin(x))²)⁶ * 2sin(x)cos(x)+1

Start by substituting t=x + (x + (sin(x)²))⁵

y=t³

Now y' = (t³)' * t'

(t³)' is easy to find. The only problem is t' =x' + ((x + (sin(x))²)⁵)'

x' is easy to find.

Now your problem is z=x + (sin(x))².

Again find the derivate using the chain rule, and go forward. After nothings left, you will go backward and find the derivative of the equation.

Regards.