Differentiate the given functions: a. f(x)=\cos(x^2e^x) b. z=\sqrt{5x+\tan(4x)}

Question:

Differentiate the given functions:

a. {eq}f(x)=\cos(x^2e^x) {/eq}

b. {eq}z=\sqrt{5x+\tan(4x)} {/eq}

Differentiation

A function with one independent variable has derivatives, we can find them if we know differentiation rules, for example, the first derivative of a constant function is zero.

a

{eq}\displaystyle \frac{d}{dx} \left[ \cos(x^2e^x) \right] \; = \; -\sin \left( x^2 {\rm e}^{x } \right)* \frac{d}{dx} \left[ x^2e^x \right] \\ \displaystyle \; = \; -\sin \left( x^2 {\rm e}^{x } \right)* \left( \frac{d}{dx} \left[ x^2 \right]* {\rm e}^{ x} \; + \; x^2 \frac{d}{dx} \left[ {\rm e}^{x } \right] \right) \\ \displaystyle \; = \; -\sin \left( x^2 {\rm e}^{x } \right)* \left( 2x*{\rm e}^{ x} \; + \; x^2 {\rm e}^{x } \right) \\ \displaystyle \; = \; -\sin \left( {x}^{2}{{\rm e}^{x}} \right) x{{\rm e}^{x}} \left( x+2 \right) \\ {/eq}

b

{eq}\displaystyle \frac{d}{dx} \left[ \sqrt{5x+\tan(4x)} \right] \; = \; \frac{ 1 }{ 2 \sqrt{ 5x+\tan(4x) } } * \frac{d}{dx} \left[ 5x+\tan(4x) \right] \\ \displaystyle \; = \; \frac{ 1 }{ 2 \sqrt{ 5x+\tan(4x) } } * \left( \frac{d}{dx} \left[ 5x \right] \; + \; \frac{d}{dx}\left[ \tan(4x) \right] \right) \\ \displaystyle \; = \; \frac{ 1 }{ 2 \sqrt{ 5x+\tan(4x) } } * \left( 5 \; + \; \sec^2(4x)* \frac{d}{dx} \left[ 4x \right] \right) \\ \displaystyle \; = \; \frac{ 1 }{ 2 \sqrt{ 5x+\tan(4x) } } * \left( 5 \; + \; \sec^2(4x)* 4 \right) \\ \displaystyle \; = \; \frac{ 5 \; + \; 4\sec^2(4x) }{ 2 \sqrt{ 5x+\tan(4x) } } \\ \displaystyle \; = \; \frac{ 1 }{2 }* \,{\frac {9+4\, \left( \tan \left( 4\,x \right) \right) ^{2}}{ \sqrt {5\,x+\tan \left( 4\,x \right) }}} \\ {/eq}