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📐 Mathematics  ·  Limits and Derivatives  ·  JEE

If y = (sin x)<sup>tan x</sup>, find dy/dx.

Answer: (sinx) tanx × [sec²x × ln(sinx) + 1].

  • A (sinx)<sup>tanx</sup> × [sec²x × ln(sinx) + 1]
  • B (sinx)<sup>tanx</sup> × sec²x, omitting the logarithmic term
  • C (sinx)<sup>tanx</sup> × ln(sinx), omitting the secant-squared term
  • D tanx × (sinx)<sup>tanx-1</sup>, treating it like a simple power rule

Correct answer: A. (sinx)<sup>tanx</sup> × [sec²x × ln(sinx) + 1]

Explanation: Take ln: y = tanx × ln(sinx). Differentiate: y'/y = sec²x × ln(sinx) + tanx × cosx/sinx = sec²x lnsinx + 1. y' = y × [sec²x lnsinx + 1].

xyP (a, f(a))Q (b, f(b))secant PQtangent at PAs Q slides toward P, secant approaches tangent

As point Q slides along the curve toward P, the secant line PQ rotates into the tangent line at P, whose slope is the derivative.

Concept context

Rate of change, differentiation rules, and applications

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