Let f be a differentiable function in (0,2).

Question

Let $f$ be a differentiable function in $\left(0,\dfrac\pi2\right)$. If $\displaystyle\int_{\cos x}^1 t^2f(t)\,dt = \sin^3x+\cos x-1$, then $\dfrac1{\sqrt3}f'\left(\dfrac1{\sqrt3}\right)$ is equal to:

Accepted Answer

Differentiating with respect to $x$ (Leibniz rule):

$$-\cos^2x\cdot f(\cos x)\cdot(-\sin x) = 3\sin^2x\cos x-\sin x$$

$$\sin x\cos^2x\,f(\cos x) = 3\sin^2x\cos x-\sin x \implies f(\cos x) = 3\tan x-\sec^2x$$

Differentiating again with respect to $x$: $-\sin x\,f'(\cos x) = 3\sec^2x-2\sec^2x\tan x$.

At $\cos x=\dfrac1{\sqrt3}$ (so $\sin x=\sqrt{2/3}$, $\tan x=\sqrt2$, $\sec^2x=3$):

$$-\sqrt{\frac23}\,f'\left(\frac1{\sqrt3}\right) = 9-6\sqrt2 \implies f'\left(\frac1{\sqrt3}\right) = -\sqrt{\frac32}(9-6\sqrt2)$$

Simplifying, $\dfrac1{\sqrt3}f'\left(\dfrac1{\sqrt3}\right) = 6-\dfrac9{\sqrt2}$.

Solutions · 1