Find Derivative: y= 1/4 x^(-4) -(cosx+3tanx)/(2sinx) -lne^(-x^2) -3e^(-7x)

[Percy]

Find first derivitave of: y= 1/4 x^(-4) -(cosx+3tanx)/(2sinx) -lne^(-x^2) -3e^(-7x)

1/4x^(-4)= -x^3

lne^(-x^2) =2x ?

-3e^(-7x) = -21e^(-7x)

Having trouble with -(cosx+3tanx)/(2sinx)

Do I use the Quotient rule?
Where do I begin?

Thanks

 

[mmm4444bot]

Find first derivitave of:

y = 1/4 x^(-4) - (cosx + 3 tanx)/(2 sinx) - lne^(-x^2) - 3e^(-7x)

d/dx [1/4x^(-4)] = -x^3

Redo that exponent calculation.


d/dx[-lne^(-x^2)] = 2x

2x is correct, with the added negative sign in front of the natural log.


d/dx[-3e^(-7x)] = -21e^(-7x)

Check your signs.


Having trouble with d/dx[-(cosx+3tanx)/(2sinx)]

Do I use the Quotient rule?
Where do I begin?

You could use the Quotient Rule. You could also use trigonometric identities to rewrite the expression, first.

You could express it as a sum of two ratios with sine and cosine, or it could be rearranged into this form:

A*cot(x) + B*sec(x)

where A and B are rational numbers. Can you find and differentiate that form? :cool:

 

[Percy]

Thanks for the help mmm4444bot! :-o

If I use the quotient rule??

y = 1/4 x^(-4) - (cosx + 3 tanx)/(2 sinx) - lne^(-x^2) - 3e^(-7x)

dy/dx=d/dx[1/4 x^(-4)] -d/dx[(cosx+3tanx)/2sinx] -d/dx [lne^(-x^2)] -d/dx [3e^(-7x)]
dy/dx=1/4 d/dx[x^(-4] -1/2 d/dx [(cosx+3tanx)/2sinx] -d/dx[inu].d/dx[e^(-x^2)] -3 d/dx[e^(-7x)]
dy/dx=1/4.-4x^(-5) - 1/2 (sinx(-sinx+3sec^(2) x) - cosx(cosx+3tanx))/sin^2 x - 1/u . e^(-x^2) . d/dx[-x^2] -3e^(-7x) . d/dx[-7x]dy/dx=-x^(-5) - (sinx(-sinx+3sec^2 x) - cosx(cosx+3tanx))/2sin^2 x - (e^(-x^2))/(e^(-x^2)) . -2x - 3e^-7x) . -7

dy/dx=-1/x^5 - (sinx(-sinx+3sec^2 x) - cosx(cosx+3tanx))/2sin^2 x -(-2x) -3(-7)e^(-7x)

dy/dx=-1/x^5 - (sinx(-sinx+3sec^2 x) - cosx(cosx+3tanx))/2sin^2 x +2x +21e^(-7x)

Don't know about (sinx(-sinx+3sec^2 x) - cosx(cosx+3tanx))/2sin



Y= Acot(x)+Bsec(x)

dy/dx= A d/dx[cotx] + B d/dx[sec(x)]dy/dx= A (-cosec^2 (x)) + B (sec(x) . tan(x))
dy/dx= -Acosec^2 (x) + B sec(x) . tan(x)

 

[mmm4444bot]

Y= Acot(x)+Bsec(x)

dy/dx= A d/dx[cotx] + B d/dx[sec(x)]

dy/dx= A (-cosec^2 (x)) + B (sec(x) . tan(x))

dy/dx= -Acosec^2 (x) + B sec(x) . tan(x)

That looks good. Were you able to find the values of A and B?