What is the derivative of f(x)=Ln(x^2*tan(x-1))?

Answer 1

f'(x) = 1/(sin(x-1)*cos(x-1)) + 2/x

f'(x) = 2/sin(2x-2) + 2/x

Answer 2

There are many steps to this question

1. Let u = -x^2tan(1-x) and use the chain rule

= - cot(1-x) / x^2 d/dx (-x^2 tan (1-x))
= cot(1-x)/ x^2 d/dx (x^2tan(1-x))

2. Use product rule where u = x^2 and v=tan(1-x)

= (cot(1-x)(x^2 d/dx (tan(1-x)) + tan(1-x) d/dx(x^2)) / x^2
= (ot(1-x)(x^2 d/dx (tan(1-x)) + 2tan(1-x)x) / x^2

3. Use the chain rule again with subs u=1-x

= (cot(1-x)(x^2 sec(1-x)^2 d/dx (1-x) + 2x tan(1-x)) / x^2
= [(cot(1-x)(x^2(d/dx(1)+d/dx(-x)) sec(1-x)^2 +2xtan(1-x))]/x^2

4. THe derivative of the constant 1 = 0

=[cot(1-x)(x^2 sec(1-x)^2 d/dx(-x)+2xtan(1-x)]/x^2
=[cot(1-x)(2xtan(1-x)-x^2 sec(1-x)^2 d/dx(x)] / x^2
= [cot(1-x)(2xtan(1-x)-x^2 sec(1-x)^2] / x^2

= 2/x - 2csc(2-2x)

The end

Answer 3

tan(x-1)+(2/x)+cot(x-1) .... or ..... 2csc(2x-2)+2x


ln(x)=(1/x)*x' ... product rule xy = x'y+y'x ... tan(x) = sec^2(x)x'
and chain rule

First the ln
1/(x^2*tan(x-1))
multiply this by (the product rule)
(2xtan(x-1)+sec^2(x-1)*1*x^2)
This equals
(2xtan(x-1)+x^2sec^2(x-1)) / (x^2tan(x-1))

substitute in a trig law

(1+tan^2(x-1)) for sec^2(x-1)

this gives you

((1+tan^2(x-1))x^2+2xtan(x-1)) / (x^2tan(x-1))

simplify

(x^2tan^2(x-1)+2xtan(x-1)+x^2) / (x^2tan(x-1))

tan(x-1)+2/x+1/tan(x-1)

tan(x-1)+2/x+cot(x+1)



You could simplify it more...



tan(x-1)+1/tan(x-1)+2/x ---->

(tan^2(x-1)+1)/tan(x-1)+2/x ---->

sec^2(x-1)/tan(x-1)+2/x ---->

1/cos^2*1/(sin(x-1)/cos(x-1)) + 2/x ---->

cos(x-1)/(sin(x-1)cos^2(x-1)) + 2/x ---->

1/(sin(x-1)cos(x-1))+2/x ---->

use double angle formula sin(x)cos(x)=(1/2)sin(2x)

1/((1/2)sin(2(x-1))+2/x ---->

2/sin(2x-2)+2x ---->

2csc(2x-2)+2x

you can use this site
http://calc101.com/webMathematica/derivatives.jsp
but you need to put in... Ln[x^(2)tan[x+1]]
and at the end change tan(x+1) for tan(x-1) it does not come out right if you put in tan(x-1)

Answer 4

we use chain rule... derivative of lnu is 1/u du...

Hence: f'(x) = [1/(x^2tan(x-1))] [(2xtan(x-1) +x^2sec^2(x-1)]
or f'(x) = 2/x + sec^2(x-1)/tan(x-1)
f'(x) = 2/x + 1/sin(x-1)cos(x-1)

Answer 5

=d/dx[ln(x^2*tan(x-1)]
=1/x^2*tan(x-1)[2xtan(x-1)+x^2sec^2(x-1)]
=(2/x)+1/(x^2sin(x-1)cos(x-1))