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Long story short, was doing a derivative worksheet given by my teacher, got a killstreak for sixty-seven questions and got stuck at No.68, now I'm pleasing for some help on a math forum to solve this so anything bad won't happen in class:

(Neither sum rule, difference rule, product rule, quotient rule, or power rule worked for me.)

\(\frac{d}{dx}\left[\tan ^5\left(\ln \left(3x^2+3\right)+8x\right)\right]\)

 

Please I'm begging you.

(This question will probably get lost in the forum section anyways.)

Jeffes02  Aug 10, 2017

Best Answer 

 #1
avatar+27052 
+3

Like so:

 

.

Alan  Aug 10, 2017
 #1
avatar+27052 
+3
Best Answer

Like so:

 

.

Alan  Aug 10, 2017
 #2
avatar
+1

Differentiate the sum term by term and factor out constants:
 = 8 d/dx(x) + d/dx(log(3 + 3 x^2)) 5 sec^2(8 x + log(3 + 3 x^2)) tan^4(8 x + log(3 + 3 x^2))


The derivative of x is 1:
 = 5 sec^2(8 x + log(3 + 3 x^2)) (d/dx(log(3 + 3 x^2)) + 1 8) tan^4(8 x + log(3 + 3 x^2))
Using the chain rule, d/dx(log(3 x^2 + 3)) = ( dlog(u))/( du) ( du)/( dx), where u = 3 x^2 + 3 and ( d)/( du)(log(u)) = 1/u:
 = 5 sec^2(8 x + log(3 + 3 x^2)) (8 + (d/dx(3 + 3 x^2))/(3 x^2 + 3)) tan^4(8 x + log(3 + 3 x^2))


Differentiate the sum term by term and factor out constants:
 = 5 sec^2(8 x + log(3 + 3 x^2)) (8 + d/dx(3) + 3 d/dx(x^2)/(3 x^2 + 3)) tan^4(8 x + log(3 + 3 x^2))


The derivative of 3 is zero:
 = 5 sec^2(8 x + log(3 + 3 x^2)) (8 + (3 (d/dx(x^2)) + 0)/(3 + 3 x^2)) tan^4(8 x + log(3 + 3 x^2))


Simplify the expression:
 = 5 (8 + (3 (d/dx(x^2)))/(3 + 3 x^2)) sec^2(8 x + log(3 + 3 x^2)) tan^4(8 x + log(3 + 3 x^2))


Use the power rule, d/dx(x^n) = n x^(n - 1), where n = 2: d/dx(x^2) = 2 x:
 = 5 sec^2(8 x + log(3 + 3 x^2)) (8 + 2 x 3/(3 + 3 x^2)) tan^4(8 x + log(3 + 3 x^2))


Simplify the expression:
Answer: | = 5 (8 + (6 x)/(3 + 3 x^2)) sec^2(8 x + log(3 + 3 x^2)) tan^4(8 x + log(3 + 3 x^2))

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