Tangent to an Exponential Curve

To find a tangent to an exponential curve, differentiate with the chain rule ($\frac{d}{dx}e^{kx} = k\,e^{kx}$), evaluate the gradient at the point, find the $y$-coordinate, and use $y - y_1 = m(x - x_1)$.

Find the equation of the tangent to the curve $y = e^{2x}$ at the point where $x = 0$. [5]

The working

Step 1, differentiate using $\frac{d}{dx}e^{2x} = 2e^{2x}$: $\frac{dy}{dx} = 2e^{2x} \quad \text{(M1, A1)}$

Step 2, gradient at $x = 0$: $m = 2e^{0} = 2(1) = 2 \quad \text{(A1)}$

Step 3, $y$-coordinate at $x = 0$ (on the curve): $y = e^{0} = 1, \quad \text{so the point is } (0, 1).$

Step 4, tangent equation with $m = 2$ through $(0, 1)$: $y - 1 = 2(x - 0) \implies y = 2x + 1 \quad \text{(A1)}$

Where the marks are won and lost

• The chain rule gives $\frac{d}{dx}e^{2x} = 2e^{2x}$: the factor of $2$ is essential. Writing just $e^{2x}$ loses the gradient.
• Read the gradient at the given $x$: $e^0 = 1$, so $m = 2$, not $2e^{2x}$.
• Find the $y$-coordinate from the curve ($y = e^0 = 1$), then use the point form.

Common mistakes

• Forgetting the chain-rule factor and using $m = e^0 = 1$.
• Taking $e^0 = 0$ instead of $1$.
• Using the gradient function $2e^{2x}$ as the tangent’s gradient without substituting $x = 0$.

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