Answers

(1) sec^2(tan[tanx])sec^2(tanx)sec^2(x)

(2) 2csc(2x)csc(4x)cot(2x) + 4csc(4x)csc(2x)cot(4x) - 20sin(5x)cos(5x)

(3) y = 4x + (4-pi)/2

(4)
a. 2cot(x)
b. (2x+3)e^(x^2+3x)
c. ln(17) 17^x
d. 5ex^(5e-1)











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Solutions

(1) Let a = tan(x) and b = tan(tan(x)) = tan a

da/dx = sec^2(x)
db/dx = sec^2(a)(da/dx)

tan(tan(tanx) = tan(b)

d/dx tan(b)
= sec^2(b)db/dx
= sec^2(b)sec^2(a)(da/dx)
= sec^2(b)sec^2(a)sec^2(x)
= sec^2(tan[tanx])sec^2(tanx)sec^2(x)


(2) Let a = -csc(4x) and b = csc(2x) and c = sin(5x)

da/dx = -(-csc(4x)cot(4x)(4)) = 4csc(4x)cot(4x)
db/dx = -csc(2x)cot(2x)(2) = -2csc(2x)cot(2x)
dc/dx = 5cos(5x)


-csc(4x)csc(2x) - 2sin^2(5x) = ab - 2c^2

d/dx = a(db/dx) + b(da/dx) - 4c(dc/dx)
= -csc(4x)[-2csc(2x)cot(2x)] + csc(2x)[4csc(4x)cot(4x)] - 4sin(5x)[5cos(5x)]
=2csc(2x)csc(4x)cot(2x) + 4csc(4x)csc(2x)cot(4x) - 20sin(5x)cos(5x)



(3) f(x) = csc(4x) + tan(2x)

f'(x) = -4csc(4x)cot(4x) + 2sec^2(2x)

f'(pi/8)
= -4csc(pi/2)cot(pi/2) + 2sec^2(pi/4)
= 0 + 2(1/[1/sqrt2])^2 = 2(2) = 4


y - 2 = f'(pi/8)(x - pi/8)
y = 4(x - pi/8) + 2 = 4x - pi/2 + 4/2 = 4x + (4-pi)/2




(4)
a. Let a = sin^2(x)

da/dx = 2sin(x)cos(x)

d/dx ln(sin^2(x)
=d/dx ln(a)
=(da/dx)/a
=[2sin(x)cos(x)]/(sin^2(x))
=2(sin(x)/sin(x))(cos(x)/sin(x))
=2(1)(cot(x)) = 2cot(x)


b. Let a = x^2 + 3x

da/dx = 2x + 3

d/dx e^(x^2+3)
=d/dx e^a
=(da/dx)e^a
=(2x+3)e^(x^2+3x)





















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