Calculus · real student question

Find dy/dx for y = x^2 / tan x.

Question

Find dydx\dfrac{dy}{dx} for

y=x2tanxy=\frac{x^2}{\tan x}

Step-by-step solution

  1. Identify the quotient rule pieces. Write y=uvy=\frac{u}{v} with u=x2u=x^2 and v=tanxv=\tan x. The quotient rule says y=uvuvv2y'=\frac{u'v-uv'}{v^2}; the order of the two products matters, so keep uvu'v first.

  2. Differentiate the numerator. By the power rule, u=2xu'=2x.

  3. Differentiate the denominator. The standard derivative of tangent is v=sec2xv'=\sec^2 x, which itself follows from tanx=sinxcosx\tan x=\frac{\sin x}{\cos x} and one application of the quotient rule.

  4. Assemble the rule. Substituting gives dydx=2xtanxx2sec2xtan2x\frac{dy}{dx}=\frac{2x\tan x-x^2\sec^2 x}{\tan^2 x}.

  5. Simplify with sine and cosine. Replacing tanx=sinxcosx\tan x=\frac{\sin x}{\cos x} and secx=1cosx\sec x=\frac{1}{\cos x} and clearing the compound fraction turns the result into 2xsinxcosxx2sin2x\frac{2x\sin x\cos x-x^2}{\sin^2 x}, which is easier to evaluate numerically.

  6. Note the domain. The formula holds wherever tanx\tan x is defined and nonzero, that is xπ2+kπx\neq\frac{\pi}{2}+k\pi and xkπx\neq k\pi.

Answer

dydx=2xtanxx2sec2xtan2x=2xsinxcosxx2sin2x\frac{dy}{dx}=\frac{2x\tan x-x^2\sec^2 x}{\tan^2 x}=\frac{2x\sin x\cos x-x^2}{\sin^2 x}

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