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Derivative of a sum of exponential, logarithmic, and power terms

Let h(x)=4ex−3ln⁡(x)+7x2.h(x) = 4e^x - 3\ln(x) + 7x^2. Find h′(x)h'(x), and then evaluate h′(1)h'(1).

Hint

Differentiate term by term using the constant multiple and sum/difference rules. Recall that ddx[ex]=ex\frac{d}{dx}[e^x] = e^x, ddx[ln⁡x]=1x\frac{d}{dx}[\ln x] = \frac{1}{x}, and use the power rule on x2x^2.

Step-by-step solution

Apply the sum/difference rule to differentiate each term separately: h′(x)=ddx[4ex]−ddx[3ln⁡x]+ddx[7x2]h'(x) = \frac{d}{dx}[4e^x] - \frac{d}{dx}[3\ln x] + \frac{d}{dx}[7x^2]. Using the constant multiple rule, pull the constants out front of each derivative.

Now use known derivative facts: ddx[ex]=ex\frac{d}{dx}[e^x] = e^x, ddx[ln⁡x]=1x\frac{d}{dx}[\ln x] = \frac{1}{x}, and the power rule gives ddx[x2]=2x\frac{d}{dx}[x^2] = 2x. Combining these gives h′(x)=4ex−3x+14xh'(x) = 4e^x - \frac{3}{x} + 14x. Substituting x=1x=1: h′(1)=4e1−31+14(1)=4e−3+14=4e+11h'(1) = 4e^1 - \frac{3}{1} + 14(1) = 4e - 3 + 14 = 4e + 11.

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