# Another Deriv

by helpm3pl3ase
Tags: deriv
 P: 79 (sec4x + 4arctanx^2)= (sec4x)(tan4x) + (4)(1/1+x^4)(2x).. Did I derive this correctly??
 Mentor P: 4,499 Did you use the chain rule on the sec(4x) term?
 P: 1,239 you forgot to multiply by $$du = 4$$ in the first term
P: 79

## Another Deriv

(sec4x)(4)(tan4x)(4) + (4)(1/1+x^4)(2x)

or

16(sec4x)(tan4x) + (4)(1/1+x^4)(2x)??
 P: 79 or do i keep just 1 4?? like this.. (sec4x)(tan4x)(4) + (4)(1/1+x^4)(2x)
 Mentor P: 4,499 It would only be one four. Let sec(4x)=sec(u). Then d(sec4x)/dx = d(secu)/dx = secu*tanu*du/dx du/dx = 4, so sec(4x)' = 4sec(4x)tan(4x)

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