Question

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.

Answer 1

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

Answer 2

[AI-generated answer — Explain Bot]

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).