if f(x) = ∫₄²ˣ cos(t²) dt, then f(x) = cos(2x²) 2 cos(4x²) 2 cos(2x²) 8x cos(4x²)
if f(x) = ∫₄²ˣ cos(t²) dt, then f(x) = cos(2x²) 2 cos(4x²) 2 cos(2x²) 8x cos(4x²)
Answer
Answer:
B. $2\cos(4x^{2})$
Explanation:
Step1: Recall the fundamental theorem of calculus and chain - rule
If $F(t)$ is an antiderivative of $\cos(t^{2})$, i.e., $F^\prime(t)=\cos(t^{2})$, then $f(x)=\int_{4}^{2x}\cos(t^{2})dt=F(2x)-F(4)$.
Step2: Differentiate $f(x)$ with respect to $x$
By the chain - rule, $f^\prime(x)=\frac{d}{dx}[F(2x)-F(4)]$. Since $F(4)$ is a constant, $\frac{d}{dx}F(4) = 0$. And $\frac{d}{dx}F(2x)=F^\prime(2x)\cdot\frac{d}{dx}(2x)$.
Step3: Calculate the derivative
We know that $F^\prime(t)=\cos(t^{2})$, so $F^\prime(2x)=\cos((2x)^{2})=\cos(4x^{2})$, and $\frac{d}{dx}(2x) = 2$. Then $f^\prime(x)=2\cos(4x^{2})$.