SACE Mathematical Methods Practice Questions

The product rule gives f′(x)=2xe^x+(x²+1)e^x (1 mark). Thus f′(0)=1 (1 mark). The point is (0,f(0))=(0,1) (1 mark). The tangent is y−1=1(x−0), or y=x+1 (1 mark).

Evaluate the point first: f(0)=(0²+1)e⁰=1, so the tangent must pass through (0,1).

Treat x²+1 and e^x as separate product factors. Their derivatives are 2x and e^x, hence f′(x)=2xe^x+(x²+1)e^x=e^x(x²+2x+1).

At x=0 the gradient is e⁰(1)=1. The point-gradient form y−1=1(x−0) keeps the point and slope visible before simplifying to y=x+1.

A common error is differentiating the product as 2xe^x. That omits the first factor multiplied by the derivative of e^x and gives the wrong gradient zero.

Substitution checks the line: x=0 gives y=1. A nearby value, x=0.01, gives f(x) close to 1.01015, consistent with initial slope about one.

The four credits are logically separate: correct product-rule derivative, gradient, point, and tangent equation. Writing only the final line hides most of that evidence.

Because the question asks for the tangent at one input, the derivative is evaluated at x=0; the variable x remains in the equation of the resulting line.

A compact independent derivation uses the local expansion e^x=1+x+O(x²) and x²+1=1+x². Their product is 1+x+2x²+O(x³), whose linear part is 1+x. That agrees with point (0,1), gradient 1 and tangent y=1+x without reusing the product-rule calculation.

The exact derivative can be factored as e^x(x+1)². This exposes two useful checks: it is nonnegative for every real x and is exactly one at zero. The tangent result therefore agrees with both the expanded product-rule form and the local behaviour of the original function.