Calculate the derivative of the function y=log2 x ln (2x) at x=1.

The Correct Answer and Explanation is:

To find the derivative of the function y=log⁡2x⋅ln⁡(2x)y = \log_2 x \cdot \ln(2x)y=log2x⋅ln(2x) at x=1x = 1x=1, we need to apply both the product rule and the chain rule.

Step 1: Apply the product rule

The product rule states that if we have two functions u(x)u(x)u(x) and v(x)v(x)v(x), then their derivative is given by:ddx[u(x)⋅v(x)]=u′(x)⋅v(x)+u(x)⋅v′(x)\frac{d}{dx} [u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x)dxd[u(x)⋅v(x)]=u′(x)⋅v(x)+u(x)⋅v′(x)

In our case, let:

u(x)=log⁡2xu(x) = \log_2 xu(x)=log2x
v(x)=ln⁡(2x)v(x) = \ln(2x)v(x)=ln(2x)

We will differentiate each of these separately.

Step 2: Differentiate u(x)=log⁡2xu(x) = \log_2 xu(x)=log2x

To differentiate u(x)=log⁡2xu(x) = \log_2 xu(x)=log2x, we can use the change of base formula, which states:log⁡2x=ln⁡xln⁡2\log_2 x = \frac{\ln x}{\ln 2}log2x=ln2lnx

Now, differentiate using the chain rule:ddxlog⁡2x=1ln⁡2⋅1x\frac{d}{dx} \log_2 x = \frac{1}{\ln 2} \cdot \frac{1}{x}dxdlog2x=ln21⋅x1

Step 3: Differentiate v(x)=ln⁡(2x)v(x) = \ln(2x)v(x)=ln(2x)

For v(x)=ln⁡(2x)v(x) = \ln(2x)v(x)=ln(2x), we apply the chain rule. First, differentiate the outer function, which is ln⁡(u)\ln(u)ln(u), where u=2xu = 2xu=2x, and then multiply by the derivative of the inner function u=2xu = 2xu=2x:ddxln⁡(2x)=12x⋅2=1x\frac{d}{dx} \ln(2x) = \frac{1}{2x} \cdot 2 = \frac{1}{x}dxdln(2x)=2×1⋅2=x1

Step 4: Apply the product rule

Now, using the product rule:y′(x)=1ln⁡2⋅1x⋅ln⁡(2x)+log⁡2x⋅1xy'(x) = \frac{1}{\ln 2} \cdot \frac{1}{x} \cdot \ln(2x) + \log_2 x \cdot \frac{1}{x}y′(x)=ln21⋅x1⋅ln(2x)+log2x⋅x1

Step 5: Evaluate at x=1x = 1x=1

Substitute x=1x = 1x=1 into the derivative:

log⁡21=0\log_2 1 = 0log21=0 because log⁡21=0\log_2 1 = 0log21=0
ln⁡(2⋅1)=ln⁡(2)\ln(2 \cdot 1) = \ln(2)ln(2⋅1)=ln(2)
11=1\frac{1}{1} = 111=1

Thus, the derivative at x=1x = 1x=1 becomes:y′(1)=1ln⁡2⋅ln⁡2+0⋅1=1y'(1) = \frac{1}{\ln 2} \cdot \ln 2 + 0 \cdot 1 = 1y′(1)=ln21⋅ln2+0⋅1=1

Final Answer:

The derivative of y=log⁡2xln⁡(2x)y = \log_2 x \ln(2x)y=log2xln(2x) at x=1x = 1x=1 is 111.