To solve, we use the product rule \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}f(x)g(x)=f'(x)g(x)+f(x)g'(x)\). Applying, we get \(\displaystyle f'(x)= (2x)'\cos(x)+2x(\cos(x))'\). Taking the derivatives we get the correct solution \(\displaystyle 2\cos(x)-2x\sin(x)\)

To find the second derivative, you must first find the first derivative. Remember to multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent:

\(\displaystyle f{}'(x)=10x-12\)

Now, take the derivative of the first derivative to find the second:

\(\displaystyle f{}''(x)=10\).

First, find the first derivative. Remember to multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent:

\(\displaystyle f{}'(x)=9x^2-16x+1\)

Now, take the second derivative:

\(\displaystyle f{}''(x)=18x-16\)

Remember that when differentiating, multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent as well.

\(\displaystyle f{}'(x)=2(\frac{5}{4})x^{\frac{1}{4}}=\frac{10}{4}x^{\frac{1}{4}}\)

Simplify to get your answer.

\(\displaystyle f{}'(x)=\frac{5}{2}x^{\frac{1}{4}}\)

To take the derivative, remember to multiply the exponent by the coefficient in front of the x term:

\(\displaystyle f{}'(x)=-6t+4\)

Before you can take the second derivative, you must take the first derivative. Remember to multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent:

\(\displaystyle f{}'(x)=12x^2-2\)

Now, take the derivative of this to find the second derivative:

\(\displaystyle f{}''(x)=24x\)

To take the derivative, remember to multiply the exponent by the coefficient and then subtract one from the exponent:

\(\displaystyle f{}'(x)=15x^2-4x+4\)

Remember that when taking the derivative, multiply the exponent by the coefficient and then subtract one from the exponent:

\(\displaystyle f{}'(x)=6x^{\frac{-1}{3}}-8x+1\)

Simplify so you don't have a negative exponent:

\(\displaystyle f{}'(x)=\frac{6}{x^{\frac{1}{3}}}-8x+1\)

To solve this problem, we use a combination of the product and chain rules. First, we use the product rule, which looks like this:

\(\displaystyle \frac{\mathrm{dy} }{\mathrm{d} x}=lnx\frac{\mathrm{d} }{\mathrm{d} x}cos(e^{x})+\frac{\mathrm{d} }{\mathrm{d} x}lnxcos(e^{x})\), which simplifies to:

\(\displaystyle \frac{dy}{dx}=lnx\frac{d}{dx}cos(e^{x})+\frac{cos(e^{x})}{x}\). We need to use the chain rule to find the derivative of \(\displaystyle cos(e^{x})\), which looks like this:

\(\displaystyle \frac{d}{dx}cos(e^{x})=-sin(e^{x})e^{x}\). Plugging this back into our equation above, we get:

\(\displaystyle \frac{dy}{dx}=-lnxsin(e^{x})e^{x}+\frac{cos(e^{x})}{x}\), which simplifies to:

\(\displaystyle \frac{dy}{dx}=\frac{cos(e^{x})}{x}-e^{x}lnxsin(e^{x})\)