Exercises
: shows the complete answer.  : gives a hint 
: plays a video solution  : shows just the final answer 
: these are important examples that illustrate new concepts, you should be sure to review the solutions to these questions 
Exercises
Find the derivatives of the following functions.


















 Find a formula for the n^{th} derivative of $f(x) = \ln x$.
 Find a formula for the n^{th} derivative of $g(x) = e^{2x}$.
 Find a formula for the n^{th} derivative of $h(x) = xe^{x}$.
Explorations
Logarithmic differentiation is a method where you start by taking the logarithm of a function, simplify it using logarithm properties then differentiate it implicitly. Thie method is particularly useful in situations where there are variables both in the base of an expression and in its exponent. Use this procedure to differentiate the following functions.






Icons courtesy of icons8.com
Find the derivative of $f(x) = \ln(x^2 + 1)$.
$$f'(x) = \frac{2x}{x^2 + 1}$$
Use implicit differentiation to find $dy/dx$ for the function $e^{\ln(x+y+1)} = x$.
Try simplifying the expression first.
Use implicit differentiation to find $dy/dx$ for the function $e^{\ln(x+y+1)} = x$.
The quick way to do this is to simplify $e^{\ln(x+y+1)} = x$ to
$$x+y+1 = x$$ $$y + 1 = 0$$ $$ y = 0$$And observe that, since this is a horizontal line, it's slope is constant at $y' = 0$.
With that said, I'll got through the implicit differentiation process so we can see that it gives us the same result.
$$e^{\ln(x+y+1)}\frac{d}{dx}(\ln(x+y+1)) = 1$$ $$e^{\ln(x+y+1)}\frac{1}{x+y+1}(1+y') = 1$$ $$(x+y+1)\frac{1}{x+y+1}(1+y') = 1$$ $$1 + y' = 1$$ $$y' = 0$$Find the derivative of $g(x) = \log_3(x^2 + 1)$.
$$g'(x) = \frac{2x}{(x^2 + 1) \ln 3}$$
Find the derivative of $f(x) = e^{x^2 + 1}$.
$$f'(x) = 2x e^{x^2 + 1}$$
Find the derivative of $h(x) = x\ln(3x+1)$.
$$h'(x) = \ln(3x+1) + \frac{3x}{3x + 1}$$
Find the derivative of $f(x) = e^{\cos x}$.
$$f'(x) = e^{\cos x}\sin x$$
Find the derivative of $f(x) = x^2 + 4^{2x1}$.
$$f'(x) = 2x + 2 \ln 4 \cdot 4^{2x1}$$
Find the derivative of $m(x) = \cos(\ln x)$.
$$m'(x) = \frac{\sin(\ln x)}{x}$$
Find the derivative of $f(x) = \frac{\ln x}{x + 1}$.
$$f(x) = \frac{x+1  x\ln x}{x(x+1)^2}$$
Find the derivative of $m(x) = \ln \frac{x+a}{xa}$.
$$m'(x) = \frac{2a(xa)}{(x+a)^3}$$
Find a formula for the n^{th} derivative of $f(x) = \ln x$.
$$f'(x) = \frac{1}{x} = x^{1}$$ $$f''(x) = x^{2}$$ $$f'''(x) = 2x^{3}$$ $$f^{(4)}(x) = 6x^{4}$$
The pattern I'm seeing is that the exponent is equal to the negative of the number of the derivative, the sign alternates between positive and negative, and the coefficient is the factorial of one less than the derivative number so
$$f^{n}(x) = (1)^{n+1}(n1)!x^{n}$$Find a formula for the n^{th} derivative of $g(x) = e^{2x}$.
$$g^{(n)}(x) = 2^n e^{2x}$$
Find a formula for the n^{th} derivative of $h(x) = xe^{x}$.
$$h'(x) = e^x + xe^x$$ $$h''(x) = e^x + (e^x + xe^x) = 2e^x + xe^x$$ $$h'''(x) = 2e^x + (e^x + xe^x) = 3e^x + xe^x$$ $$h^{(n)} = ne^x + xe^x$$
Differentiate $y = x^x$ using logarithmic differentiation.
$$y = x^x$$ $$\ln y = \ln x^x$$ $$\ln y = x \ln x$$
Now we can differentiate the function implicitly.
$$\frac{y'}{y} = x \frac{1}{x} + \ln x$$ $$\frac{y'}{y} = 1 + \ln x$$ $$y' = y(1 + \ln x)$$The last step in the process is to substitute $y$ with the original function so that the answer goes back to being strictly a function of $x$.
$$y' = x^x(1 + \ln x)$$Differentiate $y = x^{\cos x}$ using logarithmic differentiation.
$$y = x^{\cos x}$$ $$\ln y = \ln x^{\cos x}$$ $$\ln y = \cos x \ln x$$ $$\frac{y'}{y} = \sin x \ln x + \frac{\cos x}{x}$$ $$\frac{y'}{y} = \frac{x\sin x \ln x + \cos x}{x}$$ $$y' = y\frac{x\sin x \ln x + \cos x}{x}$$ $$y' = x^{\cos x}\frac{\cos x x\sin x \ln x}{x}$$
Differentiate $y = \sqrt{x}$ using logarithmic differentiation.
We could do this just using the power rule but we can also do this using logarithmic differentiate to confirm that the methods both get the same result.
Start by rewriting the root as a fractional exponent then use the $\ln x^a = a\ln x$ rule after taking the logarithm of both sides.
Differentiate $y = \sqrt{x}$ using logarithmic differentiation.
We could do this just using the power rule but we can also do this using logarithmic differentiate to confirm that the methods both get the same result.
$$y = \sqrt{x}$$ $$\ln y = \ln x^{1/2}$$ $$\ln y = \frac{1}{2}\ln x$$ $$\frac{y'}{y} = \frac{1}{2x}$$ $$y' = \frac{y}{2x}$$ $$y' = \frac{x^{1/2}}{2x}$$ $$y' = \frac{1}{2x^{1/2}}$$ $$y' = \frac{1}{2\sqrt{x}}$$Differentiate $y = (x^2 + 1)^2(x^3  1)^4$ using logarithmic differentiation.
$$y = (x^2 + 1)^2(x^3  1)^4$$ $$\ln y = \ln [(x^2 + 1)^2(x^3  1)^4]$$ $$\ln y = \ln(x^2 + 1)^2 + \ln(x^3  1)^4$$ $$\ln y = 2\ln(x^2 + 1) + 4\ln(x^3  1)$$ $$\frac{y'}{y} = \frac{4x}{x^2 + 1} + \frac{12x^2}{x^3  1}$$ $$y' = y\left(\frac{4x}{x^2 + 1} + \frac{12x^2}{x^3  1}\right)$$ $$y' = \frac{4xy}{x^2 + 1} + \frac{12x^2y}{x^3  1}$$ $$y' = \frac{4x(x^2 + 1)^2(x^3  1)^4}{x^2 + 1} + \frac{12x^2(x^2 + 1)^2(x^3  1)^4}{x^3  1}$$ $$y' = 4x(x^2 + 1)(x^3  1)^4 + 12x^2(x^2 + 1)^2(x^3  1)^3$$ $$y' = (x^2 + 1)(x^3  1)^3[4x(x^3  1) + 12x^2(x^2 + 1)$$ $$y' = (x^2 + 1)(x^3  1)^3(4x^4  4x + 12x^4 + 12x^2)$$ $$y' = (x^2 + 1)(x^3  1)^3(16x^4 + 12x^2  4x)$$