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Calculus · real student question

Differentiate and simplify the result: g(x) = e^x * ln(x^2 + 1) / (sqrt(sin x) + 2).

Question

Differentiate and simplify the result:

g(x)=exln⁡(x2+1)sin⁡x+2g(x)=\frac{e^{x}\ln\left(x^{2}+1\right)}{\sqrt{\sin x}+2}

Step-by-step solution

  1. Map the structure before differentiating anything. This is a quotient whose numerator is itself a product and whose denominator hides a chain rule. Naming the pieces keeps the bookkeeping honest:

    g(x)=u(x)v(x),u(x)=exln⁡(x2+1),v(x)=sin⁡x+2g(x)=\frac{u(x)}{v(x)},\qquad u(x)=e^{x}\ln\left(x^{2}+1\right),\qquad v(x)=\sqrt{\sin x}+2

    and the outer tool is the quotient rule

    g′(x)=u′(x)v(x)−u(x)v′(x)[v(x)]2g'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{\left[v(x)\right]^{2}}

  2. Differentiate the numerator with the product rule and an inner chain rule. With (ex)′=ex\left(e^{x}\right)'=e^{x} and (ln⁡(x2+1))′=2xx2+1\left(\ln\left(x^{2}+1\right)\right)'=\dfrac{2x}{x^{2}+1} (chain rule on ln⁡\ln of x2+1x^{2}+1):

    u′(x)=exln⁡(x2+1)+ex⋅2xx2+1=ex(ln⁡(x2+1)+2xx2+1)u'(x)=e^{x}\ln\left(x^{2}+1\right)+e^{x}\cdot\frac{2x}{x^{2}+1}=e^{x}\left(\ln\left(x^{2}+1\right)+\frac{2x}{x^{2}+1}\right)

    Factoring out exe^{x} immediately keeps the later algebra shorter.

  3. Differentiate the denominator. Write sin⁡x=(sin⁡x)1/2\sqrt{\sin x}=\left(\sin x\right)^{1/2} and apply the chain rule; the constant 22 contributes nothing:

    v′(x)=12(sin⁡x)−1/2cos⁡x=cos⁡x2sin⁡xv'(x)=\frac{1}{2}\left(\sin x\right)^{-1/2}\cos x=\frac{\cos x}{2\sqrt{\sin x}}

  4. Assemble the quotient rule. Substituting the four pieces:

    g′(x)=ex(ln⁡(x2+1)+2xx2+1)(sin⁡x+2)−exln⁡(x2+1)⋅cos⁡x2sin⁡x(sin⁡x+2)2g'(x)=\frac{e^{x}\left(\ln\left(x^{2}+1\right)+\dfrac{2x}{x^{2}+1}\right)\left(\sqrt{\sin x}+2\right)-e^{x}\ln\left(x^{2}+1\right)\cdot\dfrac{\cos x}{2\sqrt{\sin x}}}{\left(\sqrt{\sin x}+2\right)^{2}}

  5. Simplify by factoring the common exe^{x}. Both terms of the numerator carry exe^{x}, so pull it out:

    g′(x)=ex[(ln⁡(x2+1)+2xx2+1)(sin⁡x+2)−ln⁡(x2+1)cos⁡x2sin⁡x](sin⁡x+2)2g'(x)=\frac{e^{x}\left[\left(\ln\left(x^{2}+1\right)+\dfrac{2x}{x^{2}+1}\right)\left(\sqrt{\sin x}+2\right)-\dfrac{\ln\left(x^{2}+1\right)\cos x}{2\sqrt{\sin x}}\right]}{\left(\sqrt{\sin x}+2\right)^{2}}

    No further cancellation is available, because the logarithm and the radical share no factors.

  6. Note the domain and check numerically. The formula needs sin⁡x>0\sin x>0 — strictly positive, not merely non-negative, because sin⁡x\sqrt{\sin x} sits in a denominator after differentiating — so x∈(2kπ,(2k+1)π)x\in(2k\pi,(2k+1)\pi). Comparing against a central difference of gg with h=10−6h=10^{-6} at x=0.5,1,2,2.9x=0.5,1,2,2.9 gives 0.59436970.5943697, 1.51243251.5124325, 6.32524116.3252411 and 27.395139227.3951392 from both the formula and the numerical derivative, agreeing to seven significant figures.

Answer

g′(x)=ex[(ln⁡(x2+1)+2xx2+1)(sin⁡x+2)−ln⁡(x2+1)cos⁡x2sin⁡x](sin⁡x+2)2g'(x)=\frac{e^{x}\left[\left(\ln\left(x^{2}+1\right)+\frac{2x}{x^{2}+1}\right)\left(\sqrt{\sin x}+2\right)-\frac{\ln\left(x^{2}+1\right)\cos x}{2\sqrt{\sin x}}\right]}{\left(\sqrt{\sin x}+2\right)^{2}}

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