$\quad \quad \frac d{dx}e^x=e^x$
$\quad \quad \frac d{dx}b^x=b^x\ln b\;\quad \quad (b>0,\;b\neq1)$
$\quad \quad \frac d{dx}\ln x=\frac1x\;\quad \quad (x>0)$
$\quad \quad \frac d{dx}\log_bx=\left(\frac1{\ln b}\right)\left(\frac1x\right)\;\quad \quad (x>0,\;b>0,\;b\neq1)$
Find $f'\left( x \right)$ when $f(x) = 3{x^3} + 4{x^2} - 5x + 8.$
$f'(x) = 9{x^2} + 8x - 5$
Find $f'\left( x \right)$ when $f\left( x \right)\; = \;\;\ln x\; - {x^3}\;\; + 2x+e^x.$
$f'(x) = \frac{1}{x} - 3{x^2} + 2+e^x$
Find $f'\left( x \right)$ when $f\left( x \right)\; = \;\;4\ln x\; +5e^x\;\; - 7x^2.$
$f'(x) = \frac{4}{x} + 5e^x - 14x$
Find $f'\left( x \right)$ when $f\left( x \right)\; = \;\;\ln {x^8}\; - 3\ln x\;.$
$f(x)=8\ln x-3\ln x$
$f'(x) = \frac{8}{x} - \frac{3}{x}$
$f'(x) = \frac{5}{x}$
Find the equation of the tangent line to the function at the given point.
$f\left( x \right) = {e^x}\; + 2 \quad at \quad x = 0$
Point: $\quad f(0) = {e^0} + 2 = 1 + 2 = 3$
$(0,3)$
$f'(x) = {e^x}$
Slope: $\quad {m_{tan}}=f'(0) = {e^0} = 1$
${m_{tan}}=1$
$y - 3 = 1(x - 0)$
$y - 3 = 1x - 0$
$y = 1x + 3$
$y = x + 3$
$f\left( x \right) = 1\; + \ln {x^6} \quad at \quad x = e$
$f(x)=1+6lnx$
Point: $\quad f(e) = 1 + 6\ln e = 1 + 6(1) = 7$
$(e,7)$
$f(x) = 1 + 6\ln x$
$f'(x) = 0 + \frac{6}{x}$
Slope: $\quad {m_{tan}}=f'(e) = \frac{6}{e}$
${m_{tan}}= \frac{6}{e}$
$y - 7 = \frac{6}{e}(x - e)$
$y - 7 = \frac{6}{e}x - 6$
$y = \frac{6}{e}x + 1$
Applications:
The estimated resale value R (in dollars) of a company car after t years is given by $$R\left( t \right) = 24000{\left( {0.84} \right)^t}$$ What is the instantaneous rate of depreciation (in dollars per year) after:
$R'(t) = 24,000{\left( {.84} \right)^t}\;(\ln .84)$
$R'(1) = 24,000{\left( {.84} \right)^1}\;(\ln .84) = - \$3514.96/yr$
The instantaneous rate of depreciation after 1 year is $3514.96 per year.
$R'(2) = 24,000{\left( {.84} \right)^2}(\ln .84) = - \$2952.57/yr$
The instantaneous rate of depreciation after 2 years is $2952.57 per year.
$R'(3) = 24,000{\left( {.84} \right)^3}(\ln .84) = - \$2480.16/yr$
The instantaneous rate of depreciation after 3 years is $2480.16 per year.
Find the derivative of each of the following.
$f(x) = {e^x} + \ln x - 2{x^5} + 12$
$f'(x) = {e^x} + \frac{1}{x} - 10{x^4}$
$y = 7\ln x + 14x - \frac{1}{2}$
${y'} = \frac{7}{x} + 14$
$g(x) = - 5{e^x} - 12\ln x + 3{x^3}$
$g'(x) = - 5{e^x} - \frac{{12}}{x} + 9{x^2}$
$f(x) = 8\sqrt x + 7{e^x}$
$f(x) = 8{x^{1/2}} + 7{e^x}$
$f'(x) = 4{x^{ - 1/2}} + 7{e^x}$
$f'(x) = \frac{4}{{\sqrt x }} + 7{e^x}$
$y = \ln {x^7} - 3\ln x$
$y = 7\ln x - 3\ln x$
$y' = \frac{7}{x} - \frac{3}{x}$
$y' = \frac{4}{x}$
$f(x) = 4\ln \frac{1}{x} + 8$
$f(x) = 4(\ln 1 - \ln x) + 8$
$f(x) = 4\ln 1 - 4\ln x + 8$
$f'(x) = 0 - \frac{4}{x} + 0$
$f'(x) = \frac{{ - 4}}{x}$
$y = {e^x} - 7\ln 5x + 14$
$y = {e^x} - 7(\ln 5 + \ln x) + 14$
$y = {e^x} - 7\ln 5 - 7\ln x + 14$
$y' = {e^x} - 0 - \frac{7}{x} + 0$
$y' = {e^x} - \frac{7}{x}$
Find the equation of the line tangent to the graph of f at the indicated value of x.
$f(x) = 2{e^x} - 1\quad \quad at\quad x = 0$
Point: $f(0) = 2{e^0} - 1$
$f(0) = 2(1) - 1 = 1$
$(0,1)$
Slope: $f'(x) = 2{e^x}$
${m_{\tan }} = f'(0) = 2{e^0} = 2(1) = 2$
Equation of the Line: $y - 1 = 2(x - 0)$
$y - 1 = 2x$
$y = 2x + 1$
Find the equation of the line tangent to the graph of f at the indicated value of x.
$f(x) = 8\ln (x)\quad \quad at\quad x = e$
Point: $f(e) = 8\ln e$
$f(e) = 8(1) = 8$
$(e,8)$
Slope: $f'(x) = \frac{8}{x}$
${m_{\tan }}=f'(e) = \frac{8}{e}$
Equation of the Line: $y - 8 = \frac{8}{e}(x - e)$
$y - 8 = \frac{8}{e}x - 8$
$y = \frac{{8x}}{e}$
An editor of college textbooks has determined that the equation below models the sales of a calculus textbook, B (in thousands), based on the number of complimentary books sent to professors, x (also in thousands). $$B(x) = 3.24 + 1.6\ln \;(x)$$ Find and interpret the instantaneous rate of change when 6000 complimentary books are sent to professors, x=6.
$B'(x) = 0 + \frac{{1.6}}{x} = \frac{{1.6}}{x}$
$B'(6) = \frac{{1.6}}{6} = .2\overline {6} \approx .27$
When 6000 complimentary books are sent to professors, the sales of textbooks increase by 270 (.27 thousand) books.
The percentage of mothers who returned to the work force within one year after they had a child for the years 1976 through 1998 can be modeled by $$P(t) = 36.025 + 6.27\ln (t)$$ where t is years after 1977. (Source: Based on data from the Associated Press)
$t = 21\quad $when the year is 1998
$P(21) = 36.025 + 6.27\;\ln (21) = 55.11\%$
55.11% of mothers returned to the work force within one year in 1998.
$P'(21) = \frac{{6.27}}{t} = \frac{{6.27}}{{21}} = 0.30\%$
The percentage of mothers returning to work within one year was increasing at a rate of $0.30\%.$
$\frac{{P(13) - P(3)}}{{13 - 3}} = \frac{{52.11 - 42.91}}{{13 - 3}} = \frac{{9.2}}{{10}} = .92\% $
The percent of mothers returning to the work force increased on average .92% per year.
c. What happens to the rate at which the percentage is growing as more years go by?As more years go by, the change in percentage grows at a slower rate.
http://www2.fiu.edu/~rosentha/MAC2233/2233CE.htm (page 7 Section 3.3)