If y = log2 sec2 x , find the derivative dy

If y = log2 sec2 x , find the derivative dy . dx

The Correct Answer and Explanation is:

We are given the function:y=log⁡2(sec⁡2x)y = \log_2 (\sec^2 x)y=log2​(sec2x)

To find dydx\frac{dy}{dx}dxdy​, we’ll differentiate this with respect to xxx using the chain rule, properties of logarithms, and the derivatives of trigonometric functions.

Step 1: Apply the change of base formula for logarithms

We can rewrite the logarithm in base 2 as:y=ln⁡(sec⁡2x)ln⁡2y = \frac{\ln (\sec^2 x)}{\ln 2}y=ln2ln(sec2x)​

This simplifies the problem since we can now differentiate the natural logarithm of sec⁡2x\sec^2 xsec2x.

Step 2: Differentiate using the chain rule

We differentiate both sides of the equation with respect to xxx. First, apply the chain rule to differentiate ln⁡(sec⁡2x)\ln (\sec^2 x)ln(sec2x):ddx(ln⁡(sec⁡2x))=1sec⁡2x⋅ddx(sec⁡2x)\frac{d}{dx} \left( \ln (\sec^2 x) \right) = \frac{1}{\sec^2 x} \cdot \frac{d}{dx} (\sec^2 x)dxd​(ln(sec2x))=sec2x1​⋅dxd​(sec2x)

Now, differentiate sec⁡2x\sec^2 xsec2x. Using the chain rule, we know:ddx(sec⁡2x)=2sec⁡2x⋅tan⁡x\frac{d}{dx} (\sec^2 x) = 2 \sec^2 x \cdot \tan xdxd​(sec2x)=2sec2x⋅tanx

Thus, we get:ddx(ln⁡(sec⁡2x))=1sec⁡2x⋅2sec⁡2x⋅tan⁡x=2tan⁡x\frac{d}{dx} \left( \ln (\sec^2 x) \right) = \frac{1}{\sec^2 x} \cdot 2 \sec^2 x \cdot \tan x = 2 \tan xdxd​(ln(sec2x))=sec2x1​⋅2sec2x⋅tanx=2tanx

Step 3: Combine and simplify

Now, putting this into our original equation:dydx=1ln⁡2⋅2tan⁡x\frac{dy}{dx} = \frac{1}{\ln 2} \cdot 2 \tan xdxdy​=ln21​⋅2tanx

Thus, the derivative of yyy with respect to xxx is:dydx=2tan⁡xln⁡2\frac{dy}{dx} = \frac{2 \tan x}{\ln 2}dxdy​=ln22tanx​

Final Answer:

dydx=2tan⁡xln⁡2\frac{dy}{dx} = \frac{2 \tan x}{\ln 2}dxdy​=ln22tanx​

This is the derivative of y=log⁡2(sec⁡2x)y = \log_2 (\sec^2 x)y=log2​(sec2x) with respect to xxx. The key steps involved using the chain rule and properties of logarithms to simplify the problem.