Question

63-78 Find the derivative of the function. Simplify where possible.

71. \(f\left( z \right) = {e^{{\bf{arcsin}}\,\left( {{z^{\bf{2}}}} \right)}}\)

Short answer

The derivative of the function \(f\left( z \right)\) is \(\frac{{\left( {2z} \right){e^{\arcsin \left( {{z^2}} \right)}}}}{{\sqrt {1 - {z^4}} }}\).

Step-by-step solution

Write the differentiation formula for \(f\left( z \right)\)

The differentiationof \(\arcsin z\) is:

\(\frac{{\rm{d}}}{{{\rm{d}}z}}\left( {\arcsin z} \right) = \frac{1}{{\sqrt {1 - {z^2}} }}\)

The differentiation of \({e^z}\) is:

\(\frac{{\rm{d}}}{{{\rm{d}}z}}\left( {{e^z}} \right) = {e^z}\)

Find the derivative of the \(f\left( z \right)\)

The differentiation of \(f\left( z \right)\) with respect to x is:

\(\begin{aligned}f'\left( z \right) &= \frac{{\rm{d}}}{{{\rm{d}}z}}\left( {{e^{\arcsin \left( {{z^2}} \right)}}} \right)\\ &= {e^{\arcsin \left( {{z^2}} \right)}} \times \frac{{\rm{d}}}{{{\rm{d}}z}}\left( {\arcsin \left( {{z^2}} \right)} \right)\\ &= {e^{\arcsin \left( {{z^2}} \right)}}\left( {\frac{1}{{\sqrt {1 - {{\left( {{z^2}} \right)}^2}} }}} \right) \times \frac{{\rm{d}}}{{{\rm{d}}z}}\left( {{z^2}} \right)\\ &= \frac{{{e^{\arcsin \left( {{z^2}} \right)}}}}{{\sqrt {1 - {z^4}} }}\left( {2z} \right)\\ &= \frac{{\left( {2z} \right){e^{\arcsin \left( {{z^2}} \right)}}}}{{\sqrt {1 - {z^4}} }}\end{aligned}\)

Thus, the derivative of the function \(f\left( z \right)\) is \(\frac{{\left( {2z} \right){e^{\arcsin \left( {{z^2}} \right)}}}}{{\sqrt {1 - {z^4}} }}\).