The exponential function $f(x)=e^x$, where $e \approx 2.71828$, is the one such function. Using the above formula, if $f(x)=e^x$, then it must be that $f'(x)=$ . In particular, $f'(0)=$ .

If $g(z)=17 e^{z} - 9$, then $\displaystyle\diff{g}{z} =$ $\displaystyle\diff{g}{z} \biggr|_{z=1} =$

If $h(y)=5 e^{y} - y^{2}$, then $h'(y) =$ $h'(-3) =$

When substituting numbers in for the derivative, unless you happen to get $e^0$, you can't simplify the exponential. You'd just leave expressions like $e^5$ as $e^5$.

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We can use the logarithm rules to differentiate $g(y)=\ln(y^3)$. First of all, according to the rules, you can bring down the exponent to rewrite $g(y)=\ln(y^3) =$ . In that new form, $g$ can be differentiated just as above.

Calculate $g'(y)=$ and $g'(3) =$ .

If $h(t)=\ln(t^c)$ for a constant $c$, calculate $h'(t)=$ .

We can use the logarithm rules to differentiate $p(x)=\ln(6x)$. First of all, according to the rules, write the logarithm of a product as the sum of the logarithms: $p(x)=\ln(6x) =$ . In that new form, $p$ can be differentiated just as above.

Calculate $p'(x)=$ and $p'(5) =$ .

If $q(x)=\ln(cx)$ for a constant $c$, calculate $q'(x)=$ .

Use the rules derived to differentiate each term.

Online, enter the derivative $\diff{f}{x}$ as df/dx or f'(x).

$\diff{h}{x}=$ $=$ (In the first blank, enter the product rule, as in the previous part; in the second blank enter the constant product rule, i.e., the rule specialized for the case $g(x)=c$.)

Now verify you get the same answer with the product rule: $\diff{h}{x} = \diff{f}{x}g(x) + f(x)\diff{g}{x}$ $= ($$)($$)+($$)($$)$ $=$ $+$ $=$ (In second line, enter values for functions and derivatives computed above. Then, perform the multiplications in third line.)

$f'(z)=$

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First, thinking of $2t=t+t$, use the rules of exponentiation to convert $e^{t+t}$ into a product of two exponentials. $f(t) =$ $\cdot$

Each of those two exponentials in the above formula should be simple to differentiate. The product rule now makes it possible to calculate $f'(t)$. $f'(t) = $ $\cdot$ $+$ $\cdot$

To simplify the formula for $f'(t)$, use the same rule of exponentiation in the other direction, combine each term in formula for $f'(t)$ into a single exponential. $f'(t) =$ $+$ .

Lastly, combine the terms to get a simple formula for $f'(t)$. $f'(t) =$