Using the differentiation rules and the derivative table, find the derivative of the function y=4^{\sin^5(x)-3}


y' = 4^{\sin^5 x - 3}\ln(4) \cdot 5\sin^4 x \cos x

We can read the function as a composition

  • Identify the inner function and the outer function

    We write the function as

    y = 4^{g(x)}\qquad \text{con}\qquad g(x)={\sin^5(x) - 3}


    where the outer function f is the exponential with base 4

  • Apply the Chain Rule

    Remembering that the derivative of f(t)=4^t is f'(t)=4^t\ln(4), from the Chain Rule we have

    \frac{d}{dx} \left[ 4^{g(x)} \right] = 4^{g(x)}\ln(4) \cdot g'(x)


    To apply this rule to our function we need g'

  • Calculate the derivative of g


    - Derivative of \sin^5 (x):


    We write the function as a composition \sin^5 (x)= (v(x))^5 with inner function v(x)=\sin (x) and outer function (\cdot)^5. Using the Chain Rule we have

    \frac{d}{dx} [\sin^5 (x)] = 5(\sin (x))^4 \cdot \cos (x)


    - Derivative of \sin^5 (x) - 3: by linearity

    \frac{d}{dx} [\sin^5 (x) - 3] = \frac{d}{dx} [\sin^5 (x)]-\frac{d}{dx}[3]= 5\sin^4 (x) \cdot \cos (x)-0


  • Replace g and g' in the Chain Rule

    y' = 4^{\sin^5 (x) - 3}\ln(4) \cdot 5\sin^4 (x) \cos (x)