\(f\left(x\right)=\left(g\cdot h\right)\left(x\right)\)
\(g\left(x\right)=18^x\)
\(h\left(x\right)=14x^3\)
First Question is:
\(g'\left(x\right)=18x^{x-1}\)
Is that right and simplified?
Second Question is: what is \(z(x)\)
\(f''\left(x\right)=z\left(x\right)\)
Given g(x)= 18x
First Question: Find the derivative of \(g(x) = 18^x\)
\(\begin{array}{|rcll|} \hline g(x) &=& 18^x \quad & | \quad \ln() \text{ both sides } \\ \ln(~g(x)~) &=& \ln(18^x) \quad & | \quad \text{Formula: } ln(a^b) = b\cdot ln(a)\\ \ln(~g(x)~) &=& x\cdot \ln(18) \quad & | \quad \text{derivate both sides } \text{Formula: } (~ln(~g(x)~)~)' = \frac{g'(x)}{g(x)} \\ \frac{g'(x)}{g(x)} &=& ln(18) \quad & | \quad \cdot g(x) \\ g'(x) &=& g(x) \cdot ln(18) \quad & | \quad g(x) = 18^x \\\\ \mathbf{g'(x)} & \mathbf{=} & \mathbf{18^x\cdot ln(18) } \\ \hline \end{array} \)
For 2nd question:
First find f(x):
\(f(x) = (g\cdot h)(x) = g(x) \cdot h(x) = (18^x)(14x^3)\)
Then find f '(x):
In this case we use product rule:
\(f'(x) = (18^x)'(14x^3) + (18^x)(14x^3)' = 14\ln 18 \cdot 18^x \cdot x^3 +18^x\cdot42x^2\)
Let us call the part before plus sign (1) and the part after plus sign (2)
Differentiate (1):
\((1)' = 14\ln 18 \left((18^x)'(x^3)+(18^x)(x^3)'\right)\\ =14\ln 18 (x^3\cdot18^x\ln 18+3\cdot18^x\cdot x^2)\)
Differentiate (2):
\((2)' = (18^x)'(42x^2) + (18^x)(42x^2)'\\ =42\ln 18\cdot18^x\cdot x^2+84\cdot18^x\cdot x\)
Add up (1)' and (2)'
\(\quad z(x) \\= (1)' + (2)'\\ =14\ln18(x^3\cdot 18^x\ln 18+3\cdot 18^x \cdot x^2) + 42\ln 18\cdot 18^x \cdot x^2 + 84 \cdot 18^x \cdot x\\ = 14x^218^x\ln 18(x + 3) + 42x \cdot 18^x(x\ln 18+2)\)