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Differentiate?

0 votes
2) y = sec((1/2)(x)) tan ((1/2)(x))
3) y= 2cot^2((pi) (t) + 2)
4) y= cos sqrt ( sin(tan pi x))
asked Jul 22, 2014 in CALCULUS by anonymous

2 Answers

0 votes

2).

y = sec(x/2) tan(x/2).

Derivative with respect to x.

y ' = [sec(x/2) tan(x/2)] '

Apply product rule of derivative : (uv) ' = uv' + vu'.

y ' = sec(x/2)[tan(x/2)]' + tan(x/2)[sec(x/2)]'

Derivative of tan x = sec^2 x,

Derivative of sec x = sec x tan x, and

Derivative of nx with respect to x = n.

y ' = [(1/2)sec(x/2)sec^2 (x/2)] + [(1/2)tan(x/2)sec(x/2)tan(x/2)]

y ' = [(1/2)sec^3 (x/2)] + [(1/2)tan^ (x/2)sec(x/2)]

y ' = (1/2)sec(x/2)[sec^2 (x/2) + tan^ (x/2)].

3).

y = 2cot^2 (πt + 2).

Derivative with respect to t.

y ' = [2cot^2 (πt + 2)] '

y ' = 2[cot^2 (πt + 2)] '

Derivative of x^n = n * x^(n - 1).

y ' = 2[2cot (πt + 2)](cot (πt + 2)) '

Derivative of cot x = - csc^2 x.

y ' = 4cot (πt + 2)(- csc^2 (πt + 2))(πt + 2) '

Derivative of nx with respect to x = n.

y ' = 4cot (πt + 2)(- csc^2 (πt + 2))(π)

y ' = - 4πcot (πt + 2)csc^2 (πt + 2).

answered Jul 22, 2014 by lilly Expert
0 votes

4).

y = cos (√sin(tan πx)).

Differentiate the above equation with respect to x.

y ' = [ cos (√sin(tan πx)) ] '

Derivative of cos x is sin x.

y ' = sin(√sin(tan πx))(√sin(tan πx)) '

Derivative of √x is 1/(2√x).

y ' = sin(√sin(tan πx))[1/{2(√sin(tan πx))}](sin(tan πx))'

Derivative of sin x is cos x.

y ' = sin(√sin(tan πx))[1/{2(√sin(tan πx))}](cos(tan πx))(tan πx)'

Derivative of tan x is sec^2 (x).

y ' = sin(√sin(tan πx))[1/{2(√sin(tan πx))}](cos(tan πx))(sec^2 (πx))(πx)'

Derivative of nx is n.

y ' = sin(√sin(tan πx))[1/{2(√sin(tan πx))}](cos(tan πx))(sec^2 (πx))(π)

⇒ y ' = [πsin(√sin(tan πx))(cos(tan πx))(sec^2 (πx))] / [2(√sin(tan πx))].

answered Jul 23, 2014 by lilly Expert

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