# Calculus I

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• #13237

22. Find the derivative for a function given as a quotient (answer must be a simple fraction and no neg exponents!)

#16848

This was a difficult problem!!!
Okay first look at the problem and you will notice a negative exponent -3
get rid of the negative exponent by making it 1/x^3
So basically the problem should read xtanx/x2 +1/x^3
the x in the numerator should cancel with the denominator x^2
tanx/x + 1/x^3
From here basically take d/dx (tan/x) + d/dx(1/x^3)

#16852

This is what I got for my answer
= tan(x) / ( x^2+ (1/ x^3)) − ((x⋅(2x−3/x^4))⋅tan(x)) / (x^2+(1/x^3))^2 + (x⋅(sec(x))^2 / (x^2+(1/x^3))

#16855

I got xsec^2x-tanx/x^2 + -3/x^4

#16871

First step recommended is to multiply numerator and denominator by x^3 to get
x^4 tanx/(x^5+1)
Using quotient rule, we get
[5x^8tanx-(4x^3tanx +x^4sec^2x)(x^5+1)]/(x^5+1)^2

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