Question

If $y=\sin x\sin 2x$ and $t=\cos x$, then $\frac{dy}{dt}$ is:

• A. \(2(3t^2-1)\)
• B. \(1-3t^2\)
• C. \(\frac{1}{2}(1-3t^2)\)
• D. \((3t^2-1)\)
• E. \(2(1-3t^2)\)

Hint:

When one variable is given in terms of another, first rewrite the whole expression completely in that variable. Then differentiation becomes much easier.

Answer 1

Answer: (E)

Step 1: Simplify the expression for \(y\).
We are given: \[ y=\sin x \sin 2x \] Using \[ \sin 2x=2\sin x\cos x, \] we get:
\[ y=\sin x \cdot 2\sin x\cos x \] \[ y=2\sin^2 x \cos x \]

Step 2: Express everything in terms of \(t\).
Since \[ t=\cos x, \] and \[ \sin^2 x=1-\cos^2 x=1-t^2, \] we can write:
\[ y=2(1-t^2)t \]

Step 3: Expand the expression.
\[ y=2t-2t^3 \]

Step 4: Differentiate with respect to \(t\).
Now differentiate directly:
\[ \frac{dy}{dt}=\frac{d}{dt}(2t-2t^3) \]

Step 5: Compute the derivative term by term.
\[ \frac{dy}{dt}=2-6t^2 \]

Step 6: Factorize the result.
\[ \frac{dy}{dt}=2(1-3t^2) \]

Step 7: State the final answer.
Thus, \[ \boxed{\frac{dy}{dt}=2(1-3t^2)} \] which matches option \((5)\).