Since $f(x)$ is the sum of functions, remember the sum rule for derivatives.

Since several terms of $f(x)$ are a constant times a function, remember the constant multiple rule for derivatives.

Remember $x^0=1$.

Therefore,

Since $h(x)$ is the product of two functions, remember the product rule for derivatives.

Since $h(x)$ is the quotient of two functions, remember the quotient rule for derivatives.

Since $h(x)$ can be rewritten as $(3x^2-4x+2{)}^{1/2}$, remember the general power rule.

Apply the general power rule with $n=1/2$ and $f(x)=3x^2-4x+2$ and then simplify.

Since $h(x)$ is the product of two functions, remember the product rule for derivatives.

Apply the product rule with $f(x)=(4x+1{)}^3$ and $g(x)=(2x-5{)}^4$.

Notice how the general power rule was used to compute the derivative of both $(4x+1{)}^3$ and $(2x-5{)}^4$.

Since $h(x)$ is the quotient of two functions, remember the quotient rule for derivatives.

Apply the quotient rule with $f(x)=x^5$ and $g(x)=(4x-7{)}^3$.

Notice that $x=5$ because units are in thousands of calculators demanded.

In order to compute the instantaneous rate of change, we need to compute the derivative, $d^{\prime }(x)$.

Lastly, substitute $x=5$ into $d^{\prime }(x)$.

Therefore, the instantaneous rate of change of the unit price is $-20$ dollars per thousand calculators when the quantity demanded is 5000 calculators. This means that the unit price would decrease by approximately 20 dollars if the demand increased from 5000 to 6000 calculators.

Begin with the limit definition of the derivative.

Identify the $f(x)$ in the limit definition of $f^{\prime }(x)$ for our problem. If we focus on the $f(x)$ that is being subtracted in the numerator of the limit definition of $f^{\prime }(x)$, this appears to coincide with the $(x^2-5x)$ that is being subtracted in the numerator of our limit. So we hypothesize that

Verify that the given limit is equal to the limit definition of the derivative of $f(x)=x^2-5x$.

If $f(x)=x^2-5x$ then $f(x+h)=(x+h{)}^2-5(x+h)$ and therefore

Since the given limit is equal to the derivative of $x^2-5x$, we can evaluate the limit by computing the derivative of $x^2-5x$ instead of algebraically simplifying it.

Evaluate the given limit by computing the derivative of $f(x)=x^2-5x$ using the basic rules of differentiation.

Identity the $f(x)$ in the limit definition of $f^{\prime }(x)$ for our problem. If we focus on the $f(x+h)$ in the numerator of the limit definition of $f^{\prime }(x)$, this appears to coincide with the $(2+h{)}^3$ in the numerator of our limit. So we hypothesize that

Verify that the given limit is equal to the limit definition of the derivative of the function $f(x)=x^3$ at $x=2$.

If $f(x)=x^3$ then $f(2+h)=(2+h{)}^3$ and $f(2)=2^3=8$. Therefore

Since the given limit is equal to the derivative of $x^3$ at $x=2$, we can evaluate the limit by computing the derivative of $x^3$ instead of algebraically simplifying it.

Evaluate the given limit by computing the derivative of $f(x)=x^3$ and then plugging in $x=2$.