Exponential functions

The derivative of the exponential function is, by definition, itself:

ddxex=ex\displaystyle \frac{d}{dx} e^{x} = e^{x}.

With other bases, you have to multiply by the natural logarithm of the base:

ddxbx=bxlnb\displaystyle \frac{d}{dx} b^{x} = b^{x} \cdot \ln{b}.


Let’s differentiate 3x3(7)πcos2x\displaystyle 3 x^{3} \left ( 7 \right )^{\pi \cos{2} x}.

The first thing to notice is that we can use the product rule:

ddx3x3(7)πcos2x=ddx3x37πcos2x+ddx7πcos2x3x3\displaystyle \frac{d}{dx} 3 x^{3} \left ( 7 \right )^{\pi \cos{2} x} = \frac{d}{dx} 3 x^{3} \cdot 7^{\pi \cos{2} x} + \frac{d}{dx} 7^{\pi \cos{2} x} \cdot 3 x^{3},

which reduces to

9x2(7)πcos2x+7πcos2xln7ddxπcos2x\displaystyle 9 x^{2} \left ( 7 \right )^{\pi \cos{2} x} + 7^{\pi \cos{2} x} \cdot \ln{7} \cdot \frac{d}{dx} \pi \cos{2} x,

which then becomes

(9x2+ln7ddxπcos2x)7πcos2x\displaystyle \left ( 9 x^{2} + \ln{7} \frac{d}{dx} \pi \cos{2} x \right ) 7^{\pi \cos{2} x},

giving us the final answer,

(9x2ln72πsin2x)7πcos2x\displaystyle \left ( 9 x^{2} - \ln{7} \cdot 2 \pi \sin{2} x \right ) 7^{\pi \cos{2} x}.