\( y=5^x \)

\( y = x^x \)

\( y = x^x \)

Notice we could not use the power rule since the exponent is not a rational number.
Here is a video solution.

For \( f(x)=x^{2x} \), determine \(f'(x)\) and \(f'(1)\).

\( y = x^{e^x} \)

\( y = x^{e^x} \)

Notice we cannot use the power rule since the exponent, \(e^x\), is not a rational number. So to get the factor out of the exponent, we use logarithms.

\( y=(\ln x)^x \)

\( y = x^{\sin(x)} \)

\(y=(x^2)^{\sin(x)}\)

\(\displaystyle{y=\frac{(x+2)^2}{\sqrt{x^2+1}}}\)

\(\displaystyle{y=\frac{x^4\sqrt{x-3}}{(x+1)^3}}\)

\(\displaystyle{y=\sqrt{\frac{1-x}{1+x}}}\)

\( y=x^{x^2} \)

\( y=(\sin(x))^x \)

\( y=(\cos x)^{\tan x} \)

Use logarithmic differentiation to calculate \(dy/dx\) of \(\displaystyle{ y=\frac{2(x^2+1)}{\sqrt{\cos(2x)}} }\).

Use logarithmic differentiation to calculate \(dy/dx\) of \(\displaystyle{ y=\frac{2(x^2+1)}{\sqrt{\cos(2x)}} }\).

There is a mistake in this video. When he first takes the derivative, he drops a factor of two from the first term on the right side of the equal sign. Actually, I would do one more simplification than he does before taking the derivative by expanding \(\ln [2(x^2+1)] = \ln 2 + \ln(x^2+1)\). Then when you take the derivative of \(\ln 2\) you get zero.

Use logarithmic differentiation to calculate \(dy/d\theta\) of \(y=(\sin \theta)^{\sqrt{\theta}}\).

\(\displaystyle{ f(x)=(2x-3)^2(5x^2+2)^3 }\)

Use logarithmic differentiation to calculate the derivative of \(\displaystyle{ y = \frac{(\sin(x))^2(x^3+1)^4}{(x+3)^8} }\)

Use logarithmic differentiation to calculate the derivative of \(\displaystyle{ y = \sqrt{x}e^{x^2}(x^2+1)^{10} }\)

For \(\displaystyle{f(x)=\frac{x^2(x+2)^4}{(2x^2-1)^3}}\), find \(f'(x)\) and \(f'(4)\)

\(\displaystyle{y=\sqrt[3]{\frac{x(x^2+1)^4}{x^3-2}}}\)

\(\displaystyle{y=x^{(x^x)}}\)