y= (1-tan x)/sec x
find dy/dx
y= (1 + tan3x) ^(3/2)
find dy/dx
thanks a lot
y= (1-tan x)/sec x
= (1-sinx/cosx)*cosx=cosx-sinx
y'=-sinx-cosx
y=(1+tan3x)^3/2
y'=3/2(1+tan3x)^1/2 (3sec^2 3x)
for the question
y= (1 + tan3x) ^(3/2)
find dy/dx
the ans is
y'=9/2(1+tan3x)^1/2 (3sec^2 3x)
bob is right,
mrfrank has an extra 3 hanging around
should be
(9/2)(1+tan (3x))^(1/2) (sec^2 (3x) )
To find dy/dx of the given functions, we will use the rules of calculus, specifically the chain rule and quotient rule. Let's solve each problem step by step.
1. Finding dy/dx for y = (1 - tan(x))/sec(x):
Step 1: Rewrite the expression using trigonometric identities:
y = (1 - tan(x))/(1/cos(x)) = (1 - sin(x)/cos(x))/(1/cos(x)) = (1 - sin(x))/cos(x) * cos(x)/1
Step 2: Simplify the expression:
y = (1 - sin(x))/1 = 1 - sin(x)
Step 3: Differentiate both sides with respect to x:
dy/dx = d/dx (1 - sin(x))
Step 4: Apply the derivative rule:
dy/dx = 0 - cos(x) = -cos(x)
Therefore, dy/dx for y = (1 - tan(x))/sec(x) is -cos(x).
2. Finding dy/dx for y = (1 + tan(3x))^(3/2):
Step 1: Use the chain rule to differentiate the expression:
dy/dx = d/dx (1 + tan(3x))^(3/2) = (3/2)(1 + tan(3x))^(1/2) * d/dx (1 + tan(3x))
Step 2: Differentiate the inner function (1 + tan(3x)) with respect to x:
d/dx (1 + tan(3x)) = 0 + d/dx (tan(3x)) = sec^2(3x) * d/dx (3x) = 3sec^2(3x)
Step 3: Substitute back into the original expression:
dy/dx = (3/2)(1 + tan(3x))^(1/2) * 3sec^2(3x)
Step 4: Simplify the expression:
dy/dx = 9sec^2(3x) * (1 + tan(3x))^(1/2)
Therefore, dy/dx for y = (1 + tan(3x))^(3/2) is 9sec^2(3x) * (1 + tan(3x))^(1/2).
Please let me know if I can help you with anything else.