If $j(x)=e^{2x-2}+2n(x^2+x)$, find $j'(1)$

Key concept: Use the chain rule for the exponential term and the power rule for the polynomial term. Then substitute $x=1$ into the derivative.

Step by step

Step 1: Differentiate $j(x)$

Given

$j(x)=e^{2x-2}+2n(x^2+x)$

Differentiate term by term:

• For $e^{2x-2}$, use the chain rule: $\frac{d}{dx}\big(e^{2x-2}\big)=2e^{2x-2}$
• For $2n(x^2+x)$, treat $2n$ as a constant multiplier: $\frac{d}{dx}\big(2n(x^2+x)\big)=2n(2x+1)$

So,

$j'(x)=2e^{2x-2}+2n(2x+1)$

Step 2: Evaluate at $x=1$

$j'(1)=2e^{0}+2n(2\cdot 1+1)=2+6n$

Final answer

$\boxed{j'(1)=2+6n}$

Pitfall alert

A common mistake is forgetting the chain rule on $e^{2x-2}$. Another frequent error is differentiating $2n(x^2+x)$ as if $n$ were a variable; here it is treated as a constant multiplier.

Try different conditions

If the expression was intended to be $j(x)=e^{2x-2}+2\ln(x^2+x)$ instead of $2n(x^2+x)$, then the derivative would change because the logarithm needs the chain rule. In that case, you would first rewrite it carefully before evaluating at $x=1$.

Further reading

chain rule, exponential function, constant multiple rule