what is the derivative of tanx/sinx? I know the answer is secxtanx but can't figure out how to get there.

Answer 1

Tan x = sin x/cos x

So, (sin x)/(cos x) * (1/sinx) = (sin x)/ (cos x * sinx)

The two sin cancel and you are left with 1 / cos x, which is sec x
and the derivative of sec x is sec x tan x

Answer 2

Tanx Sinx

Answer 3

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RE:
what is the derivative of tanx/sinx? I know the answer is secxtanx but can't figure out how to get there.

Answer 4

We know,
tanx = sinx/cosx
Now tanx/sinx = 1/cosx = cosx^(-1) = secx.
We can use the identity that d(secx)/dx = secxtanx.
Or by using
d( ( f(x) )^n )/dx = n [(f(x)) ^(n-1)] [ d(f(x))/dx] we get the solution.
Now in this case,
taking f(x) =cosx and n = -1
Here d(f(x))/dx = d(cosx)/dx = - sinx
Now applying the above formula,
d( cosx^(-1))/dx = - cosx^(-2) * (-sinx)
= sinx/ cosx^2
= (sinx/cosx)*(1/cosx)
= tanxsecx.
and that's how we got it!

Answer 5

derivative of (tanx/sinx)
=derivative of (sinx/cosx)(1/sinx)
=derivative os i/cosx
=derivativeof secx
=secxtanx

Answer 6

Derivative of secxtanx = secxtan^2(x) + sec^3(x) = secxtan^2(x) + secx(1+tan^2(x)) = 2secxtan^2(x) + secx

Answer 7

use quoient rule: (V*(du/dx)-U*(dv/dx))/v^2 for V=sinx and U=tanx sub in and simplify

Answer 8

f(x)= tanx/sinx

f'(x)=[(sec^2x)(sinx)- (tanx)(cosx)]/(sin^2x)

f'(x)=[(sec^2x)(sinx)- sinx]/(sin^2x)

Answer 9

(tan x)/(sin x)
= {(sin x)/(cos x)}/(sin x)
= 1/(cos x)
= sec x

From first principle derivative of a function with respect to 'x' is defined as
f'(x) = Lim h->0 [{f(x+h) - f(x)}/h]
Here f(x) = sec x
So, f'(x) = Lim h->0 [{sec(x+h) - sec(x)}/h]
= Lim h->0 [{(1/cos(x+h)) - (1/cos(x))}/h]
= Lim h->0 [{cos(x) - cos(x+h)}/{h*cos(x)*cos(x+h)}]
= [1/cos(x)] * Lim h->0 [{2*sin(h/2)*sin(x + h/2)}/{h*cos(x+h)}]
= [1/cos(x)] * [Lim h->0 {2*sin(h/2)/h}] * [Lim h->0 {sin(x+h/2)/cos(x+h)}]
= [1/cos(x)] * [Lim (h/2) -->0 {sin(h/2)/(h/2)}] * [sin(x+0)/cos(x+0)]
= [1/cos(x)] * 1 * [sin(x)/cos(x)]
= sec(x)tan(x)