Register to reply 
Power and chain rule 
Share this thread: 
#1
Aug1609, 08:16 PM

P: 2

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)]^{n1} * 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 


#2
Aug1609, 09:02 PM

P: 1,622

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 [itex]u_1 = [x + (x + sin^2(x))^5][/itex] then we have that [itex]y = u_1^3[/itex]. Taking the derivative with respect to x we find that,
[tex]\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 ] [/tex] Now let [itex]u_2 = x + sin^2(x)[/itex] and try to evaluate the rest from here on out. 


#3
Aug1609, 09:22 PM

Emeritus
Sci Advisor
PF Gold
P: 16,091

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. 


#4
Aug1709, 08:25 AM

P: 365

Power and chain rule
y=t^{3} Now y' = (t^{3})' * t' (t^{3})' 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. 


Register to reply 
Related Discussions  
Using the Quotient Rule with the Chain Rule  Calculus & Beyond Homework  6  
Chain Rule/Product Rule Question  Calculus & Beyond Homework  2  
The Chain Rule, death to anyone that breaks the rule!  Calculus  55  
Combining product rule with chain rule  Calculus  2  
Chain rule in Calc = Chain in Log?  General Math  4 