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To solve the problem of finding the derivative [tex]\(\frac{dy}{dx}\)[/tex] using the chain rule, given the functions [tex]\( y = f(u) \)[/tex] and [tex]\( u = g(x) \)[/tex], specifically:
[tex]\[ y = 5u^4 - 5 \quad \text{and} \quad u = 3\sqrt{x} \][/tex]
we can break the problem down into the following steps:
1. Find [tex]\(\frac{dy}{du}\)[/tex]:
Begin by differentiating [tex]\( y \)[/tex] with respect to [tex]\( u \)[/tex]. The function [tex]\( y = 5u^4 - 5 \)[/tex] is a polynomial in terms of [tex]\( u \)[/tex].
[tex]\[ \frac{dy}{du} = \frac{d(5u^4 - 5)}{du} \][/tex]
Using the power rule of differentiation, which states that [tex]\(\frac{d}{dx}[x^n] = nx^{n-1}\)[/tex], we get:
[tex]\[ \frac{dy}{du} = 5 \cdot 4u^{3} = 20u^3 \][/tex]
2. Find [tex]\(\frac{du}{dx}\)[/tex]:
Next, we differentiate [tex]\( u \)[/tex] with respect to [tex]\( x \)[/tex]. The function [tex]\( u = 3\sqrt{x} \)[/tex] can also be written as [tex]\( u = 3x^{1/2} \)[/tex].
[tex]\[ \frac{du}{dx} = \frac{d(3x^{1/2})}{dx} \][/tex]
Again, using the power rule:
[tex]\[ \frac{du}{dx} = 3 \cdot \frac{1}{2} x^{-1/2} = \frac{3}{2} x^{-1/2} \][/tex]
Which can be simplified to:
[tex]\[ \frac{du}{dx} = \frac{3}{2\sqrt{x}} \][/tex]
3. Apply the chain rule:
The chain rule in Leibniz's notation states that:
[tex]\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \][/tex]
Substitute the expressions we found for [tex]\(\frac{dy}{du}\)[/tex] and [tex]\(\frac{du}{dx}\)[/tex] into this formula:
[tex]\[ \frac{dy}{dx} = (20u^3) \cdot \left(\frac{3}{2\sqrt{x}}\right) \][/tex]
4. Substitute [tex]\( u = 3\sqrt{x} \)[/tex] into the expression:
Since [tex]\( u = 3\sqrt{x} \)[/tex], we can substitute this back into our expression for [tex]\( \frac{dy}{dx} \)[/tex]:
[tex]\[ u = 3\sqrt{x} \implies u^3 = (3\sqrt{x})^3 = 27x^{3/2} \][/tex]
Hence:
[tex]\[ \frac{dy}{dx} = 20(27x^{3/2}) \cdot \left(\frac{3}{2\sqrt{x}}\right) \][/tex]
Simplify this expression by multiplying the coefficients and combining the terms:
[tex]\[ \frac{dy}{dx} = 20 \cdot 27x^{3/2} \cdot \frac{3}{2\sqrt{x}} \][/tex]
Combine the terms:
[tex]\[ \frac{dy}{dx} = \frac{20 \cdot 27 \cdot 3}{2} x^{(3/2) - (1/2)} \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{1620}{2} x^{1} \][/tex]
[tex]\[ \frac{dy}{dx} = 810x \][/tex]
Thus, the derivative [tex]\(\frac{dy}{dx}\)[/tex] is:
[tex]\[ \boxed{810x} \][/tex]
[tex]\[ y = 5u^4 - 5 \quad \text{and} \quad u = 3\sqrt{x} \][/tex]
we can break the problem down into the following steps:
1. Find [tex]\(\frac{dy}{du}\)[/tex]:
Begin by differentiating [tex]\( y \)[/tex] with respect to [tex]\( u \)[/tex]. The function [tex]\( y = 5u^4 - 5 \)[/tex] is a polynomial in terms of [tex]\( u \)[/tex].
[tex]\[ \frac{dy}{du} = \frac{d(5u^4 - 5)}{du} \][/tex]
Using the power rule of differentiation, which states that [tex]\(\frac{d}{dx}[x^n] = nx^{n-1}\)[/tex], we get:
[tex]\[ \frac{dy}{du} = 5 \cdot 4u^{3} = 20u^3 \][/tex]
2. Find [tex]\(\frac{du}{dx}\)[/tex]:
Next, we differentiate [tex]\( u \)[/tex] with respect to [tex]\( x \)[/tex]. The function [tex]\( u = 3\sqrt{x} \)[/tex] can also be written as [tex]\( u = 3x^{1/2} \)[/tex].
[tex]\[ \frac{du}{dx} = \frac{d(3x^{1/2})}{dx} \][/tex]
Again, using the power rule:
[tex]\[ \frac{du}{dx} = 3 \cdot \frac{1}{2} x^{-1/2} = \frac{3}{2} x^{-1/2} \][/tex]
Which can be simplified to:
[tex]\[ \frac{du}{dx} = \frac{3}{2\sqrt{x}} \][/tex]
3. Apply the chain rule:
The chain rule in Leibniz's notation states that:
[tex]\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \][/tex]
Substitute the expressions we found for [tex]\(\frac{dy}{du}\)[/tex] and [tex]\(\frac{du}{dx}\)[/tex] into this formula:
[tex]\[ \frac{dy}{dx} = (20u^3) \cdot \left(\frac{3}{2\sqrt{x}}\right) \][/tex]
4. Substitute [tex]\( u = 3\sqrt{x} \)[/tex] into the expression:
Since [tex]\( u = 3\sqrt{x} \)[/tex], we can substitute this back into our expression for [tex]\( \frac{dy}{dx} \)[/tex]:
[tex]\[ u = 3\sqrt{x} \implies u^3 = (3\sqrt{x})^3 = 27x^{3/2} \][/tex]
Hence:
[tex]\[ \frac{dy}{dx} = 20(27x^{3/2}) \cdot \left(\frac{3}{2\sqrt{x}}\right) \][/tex]
Simplify this expression by multiplying the coefficients and combining the terms:
[tex]\[ \frac{dy}{dx} = 20 \cdot 27x^{3/2} \cdot \frac{3}{2\sqrt{x}} \][/tex]
Combine the terms:
[tex]\[ \frac{dy}{dx} = \frac{20 \cdot 27 \cdot 3}{2} x^{(3/2) - (1/2)} \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{1620}{2} x^{1} \][/tex]
[tex]\[ \frac{dy}{dx} = 810x \][/tex]
Thus, the derivative [tex]\(\frac{dy}{dx}\)[/tex] is:
[tex]\[ \boxed{810x} \][/tex]
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