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Derivative of a product involving e^x and ln(x)

Let g(x)=exln⁡(x)g(x) = e^x \ln(x). Find g′(x)g'(x).

Hint

Use the product rule d/dx[f⋅h]=f′⋅h+f⋅h′d/dx[f\cdot h] = f'\cdot h + f\cdot h' with f(x)=exf(x) = e^x and h(x)=ln⁡(x)h(x) = \ln(x). Recall ddx[ex]=ex\frac{d}{dx}[e^x] = e^x and ddx[ln⁡x]=1/x\frac{d}{dx}[\ln x] = 1/x.

Step-by-step solution

Identify the two factors being multiplied: f(x)=exf(x) = e^x and h(x)=ln⁡(x)h(x) = \ln(x). Their derivatives are f′(x)=exf'(x) = e^x and h′(x)=1/xh'(x) = 1/x, using the standard transcendental derivative rules.

Apply the product rule: g′(x)=f′(x)h(x)+f(x)h′(x)=exln⁡(x)+ex⋅1xg'(x) = f'(x)h(x) + f(x)h'(x) = e^x \ln(x) + e^x \cdot \frac{1}{x}. This can be written as g′(x)=ex(ln⁡(x)+1x)g'(x) = e^x\left(\ln(x) + \frac{1}{x}\right).

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