Finding Trig. Derivatives. Please Explain Methods.?

Question

a) y = sin X tan X
b) y = sin^3 x + sec x
c) y = cos (x^2 + x + 2)
d) y = sex (1/squareroot x)

Please explain to me how I could approach this? I've read my text book and it is not helpful at all.

Also, could you possibly explain

e) Find the equation of the tangent line to the graph of y = sec(3x) at the point with x = pie/12

Thank you.

Sorry, d) is not sex it is : sec (1/ squareroot x)

Answer 1

For a) use the product rule.
y' = cos(x)tan(x) + sin(x)*(sec^2(x))
y'=sin(x)+tan(x)sec(x)

b)Use power rule:
3*sin^2(x)*(cos(x))-tan(x)sec(x)

c)use chain rule:
-sin(x^2 + x+2)*(2x+1)

d) i don't know what the derivative of sex is.

For e) find dy/dx and plug x=Pi/12 into the formula. Your answer should pop right out.

Answer 2

sinx sec^2x +tanxcosx

3sin^2 x cos x +secxtanx

-(2x+1)sin(x^2 +x+2)

Not sure what d) means

y=sec(3x)
y' = 3sec(3x)tan(3x)
When x = pi/12, y' = 3*sqrt(2)*1 = 3sqrt(2)
So y = 3xsqrt(2) +b
When x =pi/12 , y = sqrt(2)
sqrt(2) =3 pi/12 sqrt(2) +b
b = sqrt(2)- pi/4sqrt(2) =(1-pi/4)sqrt(2)
equation is y = 3xsqrt(2) + (1-pi/4)sqrt(2)