(a)
It is given that \(h\left( x \right) = {x^3}\). So, by the formula, solve as follows:
\(\begin{array}{c}h' = 3 \cdot {x^{3 - 1}} \cdot 1 + \left( {\ln x} \right) \cdot {x^3} \cdot 0\\ = 3{x^2} + 0\\ = 3{x^2}\end{array}\)
Hence, the derivative of \(h\left( x \right)\) is \(3{x^2}\).
(b)
It is given that \(h\left( x \right) = {3^x}\), So, by the formula, solve as follows:
\(\begin{array}{c}h' = x \cdot {3^{x - 1}} \cdot 0 + \left( {\ln 3} \right) \cdot {3^x} \cdot 1\\ = 0 + {3^x}\left( {\ln 3} \right)\\ = {3^x} \cdot \left( {\ln 3} \right)\end{array}\)
Hence, the derivative of \(h\left( x \right)\) is \({3^x}\left( {\ln 3} \right)\).
(c)
It is given that \(h\left( x \right) = {\left( {\sin x} \right)^x}\), So, by the formula, solve as follows:
\(\begin{array}{c}h' = x \cdot {\left( {\sin x} \right)^{x - 1}} \cdot \cos x + \left( {\ln \sin x} \right) \cdot {\left( {\sin x} \right)^x} \cdot 1\\ = x\cos x\left( {{{\sin }^{x - 1}}} \right) + {\left( {\sin x} \right)^x}\ln \left( {\sin x} \right)\\ = \frac{{x\cos x{{\left( {\sin x} \right)}^x}}}{{\sin x}} + {\left( {\sin x} \right)^x}\ln \left( {\sin x} \right)\\ = x\cot x{\left( {\sin x} \right)^x} + {\left( {\sin x} \right)^x}\ln \left( {\sin x} \right)\\ = {\left( {\sin x} \right)^x}\left( {x\cot x + \ln \sin x} \right)\end{array}\)
Hence, the derivative of \(h\left( x \right)\) is \({\left( {\sin x} \right)^x}\left( {x\cot x + \ln \sin x} \right)\).
