AHL 5.12—First principles, higher derivatives

The derivative from first principles is defined as $f^{\prime }(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$. This is the standard limit definition we must apply.

We replace $f(x)$ with $\frac{1}{x-3}$ and $f(x+h)$ with $\frac{1}{(x+h)-3}=\frac{1}{x+h-3}$. This gives $f^{\prime }(x)={\lim }_{h\to 0}\frac{\frac{1}{x+h-3}-\frac{1}{x-3}}{h}$.

The correct limit expression has both $f(x+h)$ in the numerator and $f(x)$ subtracted from it, all divided by $h$. This matches the first option exactly.

The question asks which expression correctly represents the first principles limit for $f^{\prime }(x)$ when $f(x)=\frac{1}{x-3}$. The key structure is $\frac{f(x+h)-f(x)}{h}$ inside a limit as $h\to 0$.

The second option divides by $x$ instead of $h$. The difference quotient must always be divided by the increment $h$, not by $x$, so this is incorrect.

The third option incorrectly separates the subtraction outside the limit fraction, breaking the structure of the difference quotient. It does not represent $\frac{f(x+h)-f(x)}{h}$ correctly.

The fourth option omits the $-f(x)$ term entirely from the numerator, so it is missing the subtraction of $\frac{1}{x-3}$. This does not match the definition of the derivative.

Only the first option, ${\lim }_{h\to 0}\frac{\frac{1}{x+h-3}-\frac{1}{x-3}}{h}$, correctly applies the first principles definition with $f(x+h)-f(x)$ in the numerator and $h$ in the denominator.

We write $f^{\prime }(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}={\lim }_{h\to 0}\frac{[3(x+h{)}^2+5(x+h)]-[3x^2+5x]}{h}$.

Expanding: $3(x+h{)}^2+5(x+h)=3x^2+6xh+3h^2+5x+5h$. Subtracting $f(x)=3x^2+5x$ gives the numerator $6xh+3h^2+5h$.

Factor $h$ from the numerator: ${\lim }_{h\to 0}\frac{h(6x+3h+5)}{h}={\lim }_{h\to 0}(6x+3h+5)=6x+5$.

Therefore $f^{\prime }(x)=6x+5$, which agrees with applying the power rule to each term of $3x^2+5x$.

We need to differentiate $f(x)=3x^2+5x$ from first principles. The standard power rule gives $6x+5$, so we can use that to verify.

The option $6x$ omits the derivative of $5x$, which is $5$. The constant term in the derivative of a linear term cannot be zero, so $6x$ is incorrect.

The option $3x+5$ appears to only differentiate $3x^2$ as $3x$ (forgetting to multiply by the power), so the coefficient on $x$ is wrong. It should be $6x$, not $3x$.

The option $6x+5h$ still contains $h$, which means the limit as $h\to 0$ was not properly evaluated. After taking the limit, all terms in $h$ must vanish.

The only remaining option, $6x+5$, is correct. It correctly differentiates both $3x^2\to 6x$ and $5x\to 5$.