Differentiate $y={{\tan }^{4}}\left( 2{{x}^{3}}+\ln x \right)+\cos x$ with respect to $x$. Hence

Question

Differentiate $y={{\tan }^{4}}\left( 2{{x}^{3}}+\ln x \right)+\cos x$ with respect to $x$. Hence, find $\displaystyle\int{\left( 6{{x}^{2}}+\frac{1}{x} \right){{\tan }^{3}}\left( 2{{x}^{3}}+\ln x \right){{\sec }^{2}}\left( 2{{x}^{3}}+\ln x \right)}\,\mathrm{d}x$.

Accepted Answer

(i) $4\left(6x^2+\frac1x\right)\tan^3(2x^3+\ln x)\sec^2(2x^3+\ln x)-\sin x$. (ii) $\frac14\tan^4(2x^3+\ln x)+C$.

Techniques of Differentiation - Derivatives of Trigonometric Functions

Techniques of Differentiation - Derivatives of Trigonometric Functions

Solution:

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