If Y = SecXTanX, find y" 4y' 4y.

I can easily do the second part of the question if I could if only I could find d first and second derivative of SecXTanX, pls do help.

well, finding y' and y'' is really the main part of the meal, the rest is just cleaning the dishes.

you should know the derivatives of the 6 basic trig functions by heart.

y' = secx(sec^2 x) + tanx(secxtanx) , by the product rule
= sec^3 x + tan^2 x secx

y'' = 3(sec^2 x)(secxtanx) + tan^2 x(secxtanx) + (secx)(2)(tanx)(sec^2 x)
= 3 sec^3 x tanx + secx tan^3 x + 2(sec^3 x)tanx
= 5sec^3 x tanx + secx tan^3 x

carry on

Sure! To find the first and second derivatives of Y = Sec(X)Tan(X), we'll need to use the product rule and the chain rule.

Let's begin by finding the first derivative, Y'.

Step 1: Rewrite Y = Sec(X)Tan(X) as Y = (1/Cos(X))(Sin(X)/Cos(X)).

Step 2: Simplify Y = Sin(X)/Cos^2(X).

Step 3: Apply the Quotient Rule:
Y' = (Cos^2(X)*Cos(X) - Sin(X)*(-2Cos(X)Sin(X)))/Cos^4(X).
Y' = (Cos^3(X) + 2Sin^2(X))/Cos^4(X).

Now that we have the first derivative, we can find the second derivative, Y".

Step 4: Rewrite Y' = (Cos^3(X) + 2Sin^2(X))/Cos^4(X) as Y' = [Cos^3(X)/Cos^4(X)] + [2Sin^2(X)/Cos^4(X)].

Step 5: Simplify each term separately.
a. Cos^3(X)/Cos^4(X) = Cos(X)/Cos(X)^4 = 1/Cos(X)^3.
b. 2Sin^2(X)/Cos^4(X) = 2(1-Cos^2(X))/Cos^4(X) = 2 - 2/Cos^2(X).

Therefore, Y" = 1/Cos(X)^3 - 2/Cos^2(X).

Now that we've found the first and second derivatives, let's plug them back into the original equation, Y = Sec(X)Tan(X), for the second part of the question.

Step 6: Replace Y' and Y" in the equation: 4Y' + 4Y.
4[1/Cos(X)^3 - 2/Cos^2(X)] + 4[1/Cos(X) * Sin(X)/Cos(X)].

Step 7: Simplify the equation:
4/Cos(X)^3 - 8/Cos^2(X) + 4Sin(X)/Cos^2(X).

Therefore, the expression 4y' + 4y simplifies to 4/Cos(X)^3 - 8/Cos^2(X) + 4Sin(X)/Cos^2(X).