Question

Find the derivative of the function.

48. \(y = {2^{{3^{{4^x}}}}}\)

Short answer

The derivative of the function is \(\left( {{\rm{ln}}2} \right)\left( {{\rm{ln}}3} \right)\left( {{\rm{ln}}4} \right){4^x}{3^{{4^x}}}{2^{{3^{{4^x}}}}}\), that is, \(y'\left( x \right) = \left( {{\rm{ln}}2} \right)\left( {{\rm{ln}}3} \right)\left( {{\rm{ln}}4} \right){4^x}{3^{{4^x}}}{2^{{3^{{4^x}}}}}\).

Step-by-step solution

Use the Chain rule

According to the chain rule, \(F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\), where \(F\left( x \right) = f\left( {g\left( x \right)} \right)\) and \(f\) is differentiable at \(g\left( x \right)\) and use the formula for \(\frac{d}{{dx}}\left( {{b^x}} \right) = {b^x}{\rm{ln}}b\).

Differentiate the given function with respect to \(x\) by using the chain rule:

\(\begin{aligned}y' &= \frac{d}{{dx}}\left( {{2^{{3^{{4^x}}}}}} \right)\\ &= {2^{{3^{{4^x}}}}}\left( {{\rm{ln}}2} \right)\frac{d}{{dx}}\left( {{3^{{4^x}}}} \right)\end{aligned}\)

Use the Chain rule

Again, apply the chain rule and simplify.

\(\begin{aligned}y'\left( x \right) &= {2^{{3^{{4^x}}}}}\left( {{\rm{ln}}2} \right){3^{{4^x}}}\left( {{\rm{ln}}3} \right)\frac{d}{{dx}}\left( {{4^x}} \right)\\ &= {2^{{3^{{4^x}}}}}\left( {{\rm{ln}}2} \right){3^{{4^x}}}\left( {{\rm{ln}}3} \right){4^x}\left( {{\rm{ln}}4} \right)\\ &= \left( {{\rm{ln}}2} \right)\left( {{\rm{ln}}3} \right)\left( {{\rm{ln}}4} \right){4^x}{3^{{4^x}}}{2^{{3^{{4^x}}}}}\end{aligned}\)

Thus, the derivative of the function is \(\left( {{\rm{ln}}2} \right)\left( {{\rm{ln}}3} \right)\left( {{\rm{ln}}4} \right){4^x}{3^{{4^x}}}{2^{{3^{{4^x}}}}}\), that is, \(y'\left( x \right) = \left( {{\rm{ln}}2} \right)\left( {{\rm{ln}}3} \right)\left( {{\rm{ln}}4} \right){4^x}{3^{{4^x}}}{2^{{3^{{4^x}}}}}\).