To take the derivative, remember to multiply the exponent of an x term by the coefficient in front of the x term and then subtract one from the exponent:

\(\displaystyle f{}'(x)=-27x^2-18x+4\).

Recall that the acceleration function is the second derivative of the position function. First, find the velocity function since that is the first derivative. To take the derivative, multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent:

\(\displaystyle v(t)=-32t+4\)

Now, take the derivative of the velocity function to get acceleration:

\(\displaystyle a(t)=-32\)

The first derivative of the function is equal to

\(\displaystyle f'(x)=\sec^2(2xe^x)[2e^x+2xe^x]\)

and was found using the following rule:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{d} u}{\mathrm{d} x}\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\)

The derivative of the function is equal to

\(\displaystyle f'(x)=\frac{-2(\sin(2x)-\cos(2x))}{e^{2x}}\)

and was found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)\)

The first derivative of the function is equal to

\(\displaystyle f'(x)=\frac{4x^3\cos(x)+x^4\sin(x)}{\cos^2(x)}-3\csc(3x)\cot(3x)\)

and was found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \csc(x)=-\csc(x)\cot(x)\)

The derivative of the function is equal to

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

and was found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\)

Note that the exponent is the acting as the outer function in the chain rule. 

To find the derivative, you must first use the chain rule. The derivative of \(\displaystyle e^\sin(x)\) is \(\displaystyle e^\sin(x)\), and you multiply that by the derivative of the exponent, which is \(\displaystyle \cos(x)\). Putting that all together, you get the final answer as \(\displaystyle \cos(x)e^\sin(x)\).

Taking the first derivative of the function, using the rule \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}\) and the fact that \(\displaystyle \sin'(x)=\cos(x)\), we get \(\displaystyle f'(x)= 9x^2+\cos(x)\). Using the same rule and the fact that \(\displaystyle \cos'(x)=-\sin(x)\), we get \(\displaystyle f''(x)=18x-\sin(x)\)

To find the derivative of the function, we first take the derivative of \(\displaystyle \sin(t)\) which is just \(\displaystyle \cos(t)\). Taking the derivative of the second term gets us \(\displaystyle \frac{-10}{t^3}\), using the rule \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}\), where in this case, \(\displaystyle n=-2\). Using all of the information above and applying, \(\displaystyle y'(t)= \cos(t)-\frac{10}{t^3}\)

We first take the first derivative of the function which is \(\displaystyle f'(x)=\sin(x)\). We then take the derivative of this new function, which is \(\displaystyle f''(x)=\cos(x)\). Evaluating at \(\displaystyle x=\pi/2\), we get \(\displaystyle \cos(\pi/2)=0\).