Question

If $u = \cos^3 x$ and $v = \sin^3 x$, then $\left(\frac{dv}{du}\right)_{x = \frac{\pi}{4}}$ is equal to

• A. $-2$
• B. $2$
• C. $1$
• D. $-1$

Hint:

At $x = \frac{\pi}{4}$, both $\sin x$ and $\cos x$ are equal to $\frac{1}{\sqrt{2}}$. Whenever two parametric functions have a symmetric structure with respect to sine and cosine, their derivative at $x = 45^\circ$ will almost always evaluate simply to $1$ or $-1$. This pattern saves valuable time during exams!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given two functions, $u$ and $v$, expressed parametrically in terms of a variable $x$. We need to compute the derivative of $v$ with respect to $u$, evaluated at the specific point $x = \frac{\pi}{4}$.

Step 2: Key Formula or Approach:
According to the parametric chain rule of differentiation, the derivative $\frac{dv}{du}$ is calculated by dividing their individual derivatives with respect to the parameter $x$: $$\frac{dv}{du} = \frac{\frac{dv}{dx}}{\frac{du}{dx}}$$ We will apply the power rule and trigonometric chain rules to find $\frac{du}{dx}$ and $\frac{dv}{dx}$.

Step 3: Detailed Explanation:
First, differentiate $u = \cos^3 x$ with respect to $x$: $$\frac{du}{dx} = 3\cos^2 x \cdot \frac{d}{dx}(\cos x) = 3\cos^2 x(-\sin x) = -3\cos^2 x \sin x$$ Next, differentiate $v = \sin^3 x$ with respect to $x$: $$\frac{dv}{dx} = 3\sin^2 x \cdot \frac{d}{dx}(\sin x) = 3\sin^2 x \cos x$$ Now, substitute these derivatives into the parametric chain rule formula: $$\frac{dv}{du} = \frac{3\sin^2 x \cos x}{-3\cos^2 x \sin x}$$ Cancel out the common factor $3\sin x \cos x$ from both the numerator and the denominator: $$\frac{dv}{du} = -\frac{\sin x}{\cos x} = -\tan x$$ Finally, evaluate this derivative expression at the point $x = \frac{\pi}{4}$: $$\left(\frac{dv}{du}\right)_{x = \frac{\pi}{4}} = -\tan\left(\frac{\pi}{4}\right) = -1$$ This matches option (D).

Step 4: Final Answer:
The value of the derivative is $-1$, which corresponds to option (D).