Question 2
Determine $\dfrac{dy}{dx}$ for the function $y=e^{\sin(x)}$
That’s it. Step through the hints below to see why. In 2024, 92.0% of Queensland students chose (A).
Not quite. When you differentiate $e^{f(x)}$, the power $f(x)$ stays exactly as it is. This option has changed the power from $\sin(x)$ to $\cos(x)$, so it can’t be right. Keep $e^{\sin(x)}$ and multiply it by the derivative of $\sin(x)$. In 2024, 3.8% of Queensland students chose (B).
Not quite. $e^{\sin(x)}$ is only the derivative of the outside function. The power is $\sin(x)$, not $x$, so the chain rule also multiplies by the derivative of the power, $\cos(x)$. In 2024, 2.9% of Queensland students chose (C).
Not quite. This differentiates the power but leaves the result up in the exponent. The chain rule keeps $e^{\sin(x)}$ exactly as it is and multiplies it by $\cos(x)$. In 2024, 1.2% of Queensland students chose (D).
Only (A) gives the gradient of the tangent here. Move the point and check again: (A) matches at every $x$.
The curve is falling here, so the gradient is negative. (C) and (D) are powers of $e$, which are always positive, so they can’t be right. Only (A) matches the tangent.
The tangent is flat here, so the gradient is 0. Only (A) gives 0. (C) and (D) are powers of $e$, and a power of $e$ is never 0.
Here (B) gives the same value as (A), because $\sin(x)=\cos(x)$ at this point. One point isn’t enough to decide. Move the point: only (A) keeps matching the tangent.
Here (C) gives the same value as (A), because $\cos(x)=1$. One point isn’t enough to decide. Move the point: only (A) keeps matching the tangent.
The power is $\sin(x)$, not just $x$. So this is a function inside a function.
$$y=e^{u},\qquad u=\sin(x)$$
The outside function is the exponential. The inside function is the power, $\sin(x)$.
Keep the exponential exactly as it is, then multiply by the derivative of the power.
$$\frac{d}{dx}e^{f(x)}=f'(x)\,e^{f(x)}$$
The derivative of $\sin(x)$ is $\cos(x)$, so
$$\frac{dy}{dx}=\cos(x)\,e^{\sin(x)}$$
Go to $x=\tfrac{\pi}{2}$. That is the top of the curve, so the tangent is flat and the gradient is 0.
There (A) gives $\cos\left(\tfrac{\pi}{2}\right)e^{1}=0$. (B) and (D) give 1, and (C) gives $e$. Only (A) fits.
$$\begin{aligned}\frac{dy}{dx}&=e^{\sin(x)}\times\frac{d}{dx}\big(\sin(x)\big)\\&=\cos(x)\,e^{\sin(x)}\end{aligned}$$
The exponential stays the same, and the chain rule multiplies it by the derivative of the power, $\cos(x)$.
How Queensland students answered in 2024
From the QCAA Mathematical Methods subject report for 2024. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2024 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2024, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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