How would I find dy/dx of y=(sinx *cosx *tan^3*x) / √x ?

Question

i'm trying to find the derivative

Answer 1

It requires a tedious combination of the product and chain rules. I forgot what the derivative of tan(x) is, so I can't do it for you. The product rule is the same as for only two factors, except each derivative may need to be broken up into additional instances of the product rule and/or chain rule. You needn't be bothered with the quotient rule here, because you can just multiply by x^(-0.5) instead of dividing by sqrt(x). It might also help to note that cos(x) * tan^3(x) = sin(x) * tan^2(x). If you want to check your answer, paste this into the calculator at http://www.quickmath.com/ :
(sin(x)*cos(x)*(tan(x))^3) / sqrt(x)
and differentiate. There's a link to the calculator on the side of the front page.

Answer 2

This is too lengthy to type out but heres some advice:

1. Apply the Quotient Rule to the problem as a whole.

2. As you apply the quotient rule you will need to take the derivative of the numerator which will require multiple applications of the Product Rule.

3. you will need to take the derivative Tan 3x. Apply the chain rule.

Answer 3

It is very easy...no need to confuse from others way...

first just convert sinx*cosx to (1/2)sin2x..then do it using product rule
and division rule of Differentiation

the answer is tan^3x (-2xcos2x -6x +sin2x)/(2x^2)

sorry for inablity to explain...if u need explanation, mail me

Answer 4

get a TI-89 Titanium, which will do the derivative for you. otherwise, you can do it on paper (not in your head) which will take a long time but nothing hard use the rules of derivatives like when two functions are multiplied (dy/dx of f(x)*g(x)=f'(x)*g(x)+g'(x)*f(x)) etc.

good luck thats a long one.

Answer 5

use logarithmic differentiation

lny = ln(sinx) +ln(cosx) + 3ln(tanx) -(1/2)lnx

now differentiate
(1/y)(dy/dx) = cotx -tanx +3 cotx sec^2(x) -(1/2)

(dy/dx) = y { cotx -tanx +3 cotx sec^2(x) -(1/2) }

substitute for y from the question

Answer 6

whats dy/dx

Answer 7

y=(sinxcosxtan^3x)/x^1/2=(sinxcosxsin^3x/cos^x)/x^1/2=(sin^4x/cos^2x)/x^1/2 now taking log logy=4logsinx-2logcosx-1/2logx and differentiating1/y*dy/dx=4cosx/sinx+2sinx/cosx
-1/(2x) therefore dy/dx=y[8xcos^2x+2xsin^2x-sinxcosx]
/[2xsinxcosx]=[8xsinxcos^3xtan^3x+2xsin^3xcosxtan^3x- sin^2xcos^2xtan^3x]/2x^1/2sinxcosx
(4x^1/2sin^2xtanx)+x^1/2sin^2xtan^3x)-(1/2 cot^2x)