How do you differentiate #G(x) = intsqrtt sint dt# from #sqrt(x)# to #x^3#?

Question

How do you differentiate #G(x) = intsqrtt sint dt#  from #sqrt(x)# to #x^3#?

Isaiah Allen · Accepted Answer

You have

#G(x)=int_sqrt(x)^(x^3)sqrt(t)*sintdt=>d(G(x))/dx=sqrt(x^3)*sin(x^3)*(3x^2)-sqrt(sqrt(x))*sin(sqrt(x))*(1/(2*sqrt(x)))#

Luke Alvarez · Answer

undefined To differentiate $ G(x) = \int_{\sqrt{t}}^{\sin(t)} dt $ from $ \sqrt{x} $ to $ x^3 $, we first need to differentiate the integral expression with respect to $ x $ using the Fundamental Theorem of Calculus. Then, we differentiate $ \sqrt{x} $ and $ x^3 $ separately with respect to $ x $. Let's denote the integral function as $ F(x) $.

The Fundamental Theorem of Calculus states that if $ F(x) = \int_{a(x)}^{b(x)} f(t) dt $, then $ F'(x) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x) $.

Differentiating $ F(x) $, we get:

$$ F'(x) = \sin(\sin(x)) \cdot \frac{d}{dx}(\sin(x)) - \sin(\sqrt{x}) \cdot \frac{d}{dx}(\sqrt{x}) $$

For $ \sqrt{x} $, the derivative is $ \frac{1}{2\sqrt{x}} $.

For $ x^3 $, the derivative is $ 3x^2 $.

Therefore, we have:

$$ F'(x) = \sin(\sin(x)) \cdot \cos(x) - \frac{\sin(\sqrt{x})}{2\sqrt{x}} $$

To differentiate $ \sqrt{x} $, we simply apply the power rule, which gives us $ \frac{1}{2\sqrt{x}} $.

To differentiate $ x^3 $, we apply the power rule again, which gives us $ 3x^2 $.

Thus, we differentiate $ G(x) = \int_{\sqrt{t}}^{\sin(t)} dt $ with respect to $ x $ to get $ \sin(\sin(x)) \cdot \cos(x) - \frac{\sin(\sqrt{x})}{2\sqrt{x}} $.

The differentiation of $ \sqrt{x} $ with respect to $ x $ is $ \frac{1}{2\sqrt{x}} $, and the differentiation of $ x^3 $ with respect to $ x $ is $ 3x^2 $.