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\(f\left(x\right)=\left(g\cdot h\right)\left(x\right)\)

\(g\left(x\right)=18^x\)

\(h\left(x\right)=14x^3\)

 

First Question is:

\(g'\left(x\right)=18x^{x-1}\)

Is that right and simplified?

 

Second Question is: what is \(z(x)\)

\(f''\left(x\right)=z\left(x\right)\)

 Nov 24, 2016
 #1
avatar+26387 
0

Given g(x)= 18x

 

First Question: Find the derivative of  \(g(x) = 18^x\)

\(\begin{array}{|rcll|} \hline g(x) &=& 18^x \quad & | \quad \ln() \text{ both sides } \\ \ln(~g(x)~) &=& \ln(18^x) \quad & | \quad \text{Formula: } ln(a^b) = b\cdot ln(a)\\ \ln(~g(x)~) &=& x\cdot \ln(18) \quad & | \quad \text{derivate both sides } \text{Formula: } (~ln(~g(x)~)~)' = \frac{g'(x)}{g(x)} \\ \frac{g'(x)}{g(x)} &=& ln(18) \quad & | \quad \cdot g(x) \\ g'(x) &=& g(x) \cdot ln(18) \quad & | \quad g(x) = 18^x \\\\ \mathbf{g'(x)} & \mathbf{=} & \mathbf{18^x\cdot ln(18) } \\ \hline \end{array} \)

 

 

 laugh

 Nov 24, 2016
edited by heureka  Nov 24, 2016
 #2
avatar+9665 
0

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)'

\(\quad z(x) \\= (1)' + (2)'\\ =14\ln18(x^3\cdot 18^x\ln 18+3\cdot 18^x \cdot x^2) + 42\ln 18\cdot 18^x \cdot x^2 + 84 \cdot 18^x \cdot x\\ = 14x^218^x\ln 18(x + 3) + 42x \cdot 18^x(x\ln 18+2)\)

 Nov 24, 2016

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