Exercises

Approach: $F(x)=\int_0^x t^2\,dt$ is an upper-limit integral, so the Fundamental Theorem of Calculus gives $F'(x)=f(x)$, where $f$ is the integrand.

Steps:

1. Identify the integrand: $f(t)=t^2$, so $f(x)=x^2$.
2. Apply the theorem: $F'(x)=f(x)$ for the upper-limit integral $F(x)=\int_0^x f(t)\,dt$.
3. Substitute: $F'(x)=x^2$.

Answer: $F'(x) = x^2$.

Approach: $F(x)=\int_1^x \dfrac{1}{t^2}\,dt$ is an upper-limit integral with a constant lower limit, so the derivative of $F$ is the integrand evaluated at the upper limit.

Steps:

1. Identify the integrand: $f(t)=\dfrac{1}{t^2}$, so $f(x)=\dfrac{1}{x^2}$.
2. Apply the Fundamental Theorem of Calculus: $F'(x)=f(x)$.
3. Substitute: $F'(x)=\dfrac{1}{x^2}$.

Answer: $F'(x) = \dfrac{1}{x^2}$.

Approach: The upper limit $x^3$ is itself a function of $x$, so differentiate the upper-limit integral and multiply by the derivative of the upper limit.

Steps:

1. Identify the integrand: $f(t)=\cos t$, so $f(x^3)=\cos(x^3)$.
2. Apply the chain rule to the upper limit: $\dfrac{d}{dx}\int_0^{x^3} f(t)\,dt = f(x^3)\cdot(x^3)'$.
3. Compute $(x^3)'=3x^2$ and substitute.

Answer: $F'(x) = 3x^2 \cos(x^3)$.

Approach: Differentiate the upper-limit integral once to get $f'(x)$, differentiate again to get $f''(x)$, and then apply the second derivative test at $x=0$.

Steps:

1. First derivative: $f'(x)=e^{-x^2}\cos x$, so $f'(0)=e^0\cos 0=1$.
2. Second derivative: $f''(x)=(-2x)e^{-x^2}\cos x - e^{-x^2}\sin x$, so $f''(0)=0$.
3. Test $x=0$: because $f''(0)=0$, the second derivative test is inconclusive and $f'$ does not change sign near $0$.

Answer: $f'(0)=1$, $f''(0)=0$, so $x=0$ is not an extremum.

Approach: The given limit fixes the behaviour of the integrand near $x=1$, but the derivative of the upper-limit integral is obtained directly from the integrand by the Fundamental Theorem of Calculus.

Steps:

1. Identify the integrand: for $f(x)=\int_0^x e^{t^2}\sin t\,dt$ the integrand is $g(t)=e^{t^2}\sin t$.
2. Apply the theorem: $f'(x)=g(x)=e^{x^2}\sin x$.
3. Evaluate at $x=1$: $f'(1)=e^{1}\sin 1=e\sin 1$.

Answer: $f'(1) = e \sin 1$.

Approach: The parametric equations describe the curve through the origin, while the derivative of the integral $f(x)=\int_0^x e^{t^2}\sin t\,dt$ is obtained directly from its integrand.

Steps:

1. From the parametric equations, $t=0$ gives $x=0$ and $y=0$, so the curve passes through the origin.
2. Identify the integrand: $g(t)=e^{t^2}\sin t$.
3. Apply the theorem: $f'(x)=g(x)=e^{x^2}\sin x$, so $f'(0)=e^0\sin 0=0$.

Answer: $f'(0) = 0$.

Approach: Differentiate the upper-limit integral to obtain the integrand, then differentiate that expression once more with the chain rule.

Steps:

1. First derivative: $f'(x)=e^{\cos x}$.
2. Second derivative: apply the chain rule to $e^{\cos x}$, so $f''(x)=e^{\cos x}\cdot(-\sin x)=-e^{\cos x}\sin x$.

Answer: $f'(x)=e^{\cos x}$, $f''(x)=-e^{\cos x}\sin x$.

Approach: Differentiate the upper-limit integral to obtain the integrand, then differentiate the product $e^{x^2}\sin x$ with the product rule.

Steps:

1. First derivative: $f'(x)=e^{x^2}\sin x$.
2. Product rule: $f''(x)=(e^{x^2})'\sin x + e^{x^2}(\sin x)' = 2x e^{x^2}\sin x + e^{x^2}\cos x$.

Answer: $f'(x) = e^{x^2} \sin x$, $f''(x) = 2x e^{x^2} \sin x + e^{x^2} \cos x$.

Approach: The differential equation constrains the function $y$, but the derivative that is asked for belongs to the integral $f(x)=\int_0^x e^{t^2}\sin t\,dt$, whose derivative is its integrand.

Steps:

1. Identify the integrand: $g(t)=e^{t^2}\sin t$.
2. Apply the Fundamental Theorem of Calculus: $f'(x)=g(x)=e^{x^2}\sin x$.
3. Evaluate at $x=1$: $f'(1)=e^{1}\sin 1=e\sin 1$.

Answer: $f'(1) = e \sin 1$.

Approach: Differentiate the upper-limit integral twice and evaluate at $x=0$; the conditions on $f$ guarantee that the integrals involved are well defined.

Steps:

1. First derivative: $F'(x)=e^{x^2}\sin x$, so $F'(0)=e^0\sin 0=0$.
2. Second derivative: $F''(x)=(e^{x^2})'\sin x + e^{x^2}(\sin x)' = 2x e^{x^2}\sin x + e^{x^2}\cos x$.
3. Evaluate at $x=0$: $F''(0)=0 + e^0\cos 0=1$.

Answer: $F'(0) = 0$, $F''(0) = 1$.