Question

Differentiate \(log⁡(cos⁡(sin⁡(e^{x^3})))\) w.r.t x.

(a) –\(3x^2 \,e^{x^3} \,cos⁡e^{x^3} \,tan⁡(sin⁡e^{x^3})\)

(b) \(3x^2 \,e^{x^3} \,cos⁡e^{x^3} \,tan⁡(sin⁡e^{x^3})\)

(c) –\(3e^{x^3} \,cos⁡e^{x^3} \,cos⁡(sin⁡e^{x^3})\)

(d) –\(x^2 e^{x^3} \,cos⁡e^{x^3} \,tan⁡(sin⁡e^{x^3})\)

The question was asked in an online quiz.

Asked question is from Exponential and Logarithmic Functions in portion Continuity and Differentiability of Mathematics – Class 12

Answer 1

Correct option is (a) –\(3x^2 \,e^{x^3} \,cos⁡e^{x^3} \,tan⁡(sin⁡e^{x^3})\)

Best explanation: Consider y=\(log⁡(cos⁡(sin⁡(e^{x^3})))\)

Differentiating w.r.t x by using chain rule, we get

\(\frac{dy}{dx}\)=\(\frac{d}{dx} (log⁡(cos⁡(sin⁡(e^{x^3}))))\)

\(\frac{dy}{dx}\)=\((\frac{1}{cos⁡(sin⁡e^{x^3})} \frac{d}{dx} (cos⁡(sin⁡e^{x^3})))\)

\(\frac{dy}{dx}\)=\((\frac{1}{cos⁡(sin⁡e^{x^3})} (-sin⁡(sin⁡e^{x^3}) \frac{d}{dx}(sin⁡e^{x^3})))\)

\(\frac{dy}{dx}\)=\((\frac{1}{cos⁡(sin⁡e^{x^3})} (-sin⁡(sin⁡e^{x^3})(cos⁡e^{x^3}) \frac{d}{dx} (e^{x^3})))\)

\(\frac{dy}{dx}\)=\((\frac{1}{cos⁡(sin⁡e^{x^3})} (-sin⁡(sin⁡e^{x^3})(cos⁡e^{x^3})(e^{x^3}) \frac{d}{dx} {x^3}))\)

\(\frac{dy}{dx}\)=\((\frac{1}{cos⁡(sin⁡e^{x^3})} (-sin⁡(sin⁡e^{x^3}).cos⁡e^{x^3} .e^{x^3}.3x^2)\)

\(\frac{dy}{dx}\)=-\((\frac{3x^2 e^{x^3} cos⁡e^{x^3} sin⁡(sin⁡e^{x^3})}{cos⁡(sin⁡e^{x^3})})\)

\(\frac{dy}{dx}\)=-\(3x^2 e^{x^3} cos⁡e^{x^3} tan⁡(sin⁡e^{x^3})\)