Calculus
posted by Anonymous .
1.) Find the derivative of tan (sec x).
2.) Find the derivative if 1/x in four ways, using the limit process, power rule, quotient rule and implicit differentiation.
3.) Show that the derivative of sec^1 x is 1/(x*sqrt(x^2 1)).
4. Find the derivative of 2^(e^(sin x)).
Thank you in advance.

Calculus 
Steve
d/du(tan u) = sec^2 u
so, use that and apply the chain rule:
sec^2(sec x) * secx tanx
I'm sure you can do #2
y = sec^1 x
sec y = x
secy tany y' = 1
y' = 1/(secy tany)
y' = 1/(x √(x^21))
y = 2^(e^(sinx))
y' = ln2 2^(e^(sinx)) * e^(sinx) * cosx
Respond to this Question
Similar Questions

Math  Calculus Question.
hey can someone explain to me the relationship between the chain rule and implicit differentiation? 
calculus
find dy/dx y=ln (secx + tanx) Let u= secx + tan x dy/dx= 1/u * du/dx now, put the derivative of d secx/dx + dtanx/dx in. You may have some challenging algebra to simplify it. Use the chain rule. Let y(u) = ln u u(x) = sec x + tan x … 
calculus pleas pleas help, derivates
what is the derivative of x/tan(x) and find f'(pi/4) f'(x)=(tanxx*sec(x)^2)/(tanx)^2 That's the quotient rule for derivatives or as my teacher says: Low,low, low D'High, minus High d'low all over low square that's the quotient rule. … 
calculus
Alright so implicit differentiation is just not working out for me. Use implicit differentiation to find the slope of the tangent line to the curve at point (4,1). y / (x2y) = x^3 + 4 Tried quotient rule to get the derivative of the … 
12th Grade Calculus
Find d^2y/dx^2 by implicit differentiation. x^(1/3) + y^(1/3) = 4 I know that first you must find the 1st derivative & for y prime I got 1/3x^(2/3) + 1/3y^(2/3) dy/dx = 0 Then for dy/dx I got dy/dx = [1/3x^(2/3)] / [1/3y^(2/3)] … 
calculus
Please help. Applying the chain rule, how do I find the derivative of f(x)=In(e^xe^x) (x>0) and then using this answer use the quotient rule to find the second derivative. I cannot do this so any answer greatly appreciated. 
calculus
Given y^4 x^4=16 find and simplify d2y/dx2 using implicit differentiation. I got the first derivative which was x^3/y^3 and no im am stuck on using implicit differentiation for the second derivative part 
applied calculus
1. Consider the function y = xx (for x> 0). a) Why does the derivative rule for xn not apply? 
Calculus
Differentiate to find critical numbers and leave function in fully factored form. g(x) = (x^2+1)^5(x^2+2)^6 g'(x) = (x^2+1)^5[6(x^2+2)^5(2x)] + (x^2+2)^6[5(x^2+1)^4(2x)] g'(x) = 2x(x^2+1)^4(x^2+2)^5[6(x^2+2)(2x) + 5(x^2+1)(2x)] g'(x) … 
Calculus
Find the derivative of [(2+x)/(x3)]^(2/5) I tried the power of a function rule, quotient rule, the chain rule but keep getting stuck.