Answer:
h'(x)= [tex]e^{x^{9}+In(x) } (9x^{8} +\frac{1}{x} )[/tex]
Step-by-step explanation:
h(x)=[tex]e^{x^{9} +In(x)}[/tex]
first, we will use the chain rule
step1: we divide the function into two parts So,
f(x)=[tex]e^{x^{9} +In(x)}[/tex] and g(x)=[tex]x^{9} +In(x)[/tex]
step2: rewrite [tex]x^{9} +In(x)[/tex] as u.
f(x)=[tex]e^{u}[/tex] and g(x)=u
step3: differentiate f(x) first.
f'(x)=[tex]e^{u}[/tex] (the formula is y=[tex]e^{x}[/tex] and y'=[tex]e^{x}[/tex] - the same)
step4: write the value of u for each function but now sub it in the differentiable form of f(x).
f'(x)=[tex]e^{x^{9}+In(x) }[/tex] and g(x)=[tex]x^{9}+In(x)[/tex]
step5: differentiate the function g(x).
g'(x)=[tex]9x^{8} +\frac{1}{x}[/tex]
(for [tex]x^{9}[/tex] the formula is y=[tex]x^{n}[/tex] and y'=[tex]nx^{n-1}[/tex] ) (for In(x) the formula is y=In(x) and y'=[tex]\frac{1}{x}[/tex] )
step6: use the chain rule to finish out So,
h'(x)=f'(x) × g'(x)
=[tex]e^{x^{9}+In(x) }[/tex] × [tex]9x^{8}+\frac{1}{x}[/tex]
step7: simplify.
h'(x)= [tex]e^{x^{9} +In(x)}(9x^{8}+\frac{1}{x})[/tex]
