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Sagot :
Answer:
y' = 7 * (e^(7x) * (x + 1))/((x + 2)^8) = (e^x/(x + 2))^7(7 - 7/(x + 2))
Step-by-step explanation:
The Chain Rule
The chain rule claims that for any composite function, h, composed by two differentiable functions, f and g, the following is always true:
[tex]h(x) = (f\circ g)(x)\\\\\implies h'(x)=(f'\circ g)(x)\cdot g'(x)[/tex]
The Quotient Rule
The quotient rule claims that for any function, h = (f(x))/(g(x)), the following is always true for g(x) ≠ 0:
[tex]h(x)=\frac{f(x)}{g(x)}\\\\h'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{\left(g(x)\right)^2}[/tex]
Working out the Derivative
We need to find the derivative of the following function:
[tex]y=\left(\frac{e^x}{x+2}\right)^7[/tex]
let f(x) = x^7 and g(x) = (e^x)/(x + 2), then y = f(g(x)), which we can differentiate using the chain rule:
[tex]y' = f'(g(x))\times g'(x)=7\left(g(x)\right)^6\times g'(x)=7\left(\frac{e^x}{x+2}\right)^6\times\left[\frac{e^x}{x+2}\right]'[/tex]
We'll find the derivative of g(x) using the quotient rule:
[tex]\left[\frac{e^x}{x+2}\right]'=\frac{[e^x]'(x+2)-e^x[x+2]'}{\left(x+2\right)^2}=\frac{e^x(x+2)-e^x\times1}{(x + 2)^2}=\frac{e^x\left(x+1\right)}{\left(x+2\right)^2}[/tex]
Substituting back into the equation of y:
[tex]y' = 7\left(\frac{e^x}{x+2}\right)^6\times\frac{e^x(x+1)}{(x + 2)^2}[/tex]
We can simplify this derivative to be in the requested form:
[tex]y'=7\times\frac{e^{6x}}{(x+2)^6}\times\frac{e^x(x+1)}{(x+2)^2}\\\\\to y'=\frac{e^{7x}}{(x+2)^7}\times\frac{7(x + 1)}{x+2}\\\\\to y'=\left(\frac{e^x}{x+2}\right)^7\times\frac{7x+7}{x+2}\\\\\to y'=\left(\frac{e^x}{x+2}\right)^7\times\frac{7x + 7 + 7 - 7}{x + 2}\\\\\to y'=\left(\frac{e^x}{x+2}\right)^7\times\frac{7(x + 2) - 7}{x + 2}\\\\\to y'=\left(\frac{e^x}{x+2}\right)^7\left(7-\frac7{x+2}\right)[/tex]
The derivative of y = (e^x/x + 2)^7 is y' = (e^x/(x + 2))^7(7 - 7/(x + 2))
Answer:
[tex]\dfrac{\text{d}y}{\text{d}x}=\left(\dfrac{e^x}{x+2}\right)^7\left(7 -\dfrac{7}{x+2}\right)[/tex]
Step-by-step explanation:
Given equation:
[tex]y=\left(\dfrac{e^x}{x+2}\right)^7[/tex]
To differentiate the equation, we can use the chain rule.
[tex]\boxed{\begin{array}{c}\underline{\text{Chain Rule for Differentiation}}\\\\\text{If\;\;$y=f(u)$\;\;and\;\;$u=g(x)$\;\;then:}\\\\\dfrac{\text{d}y}{\text{d}x}=\dfrac{\text{d}y}{\text{d}u}\times\dfrac{\text{d}u}{\text{d}x}\end{array}}[/tex]
[tex]\textsf{Let $u$}=\dfrac{e^x}{x+2}[/tex]
Rewrite y in terms of u:
[tex]y=u^7[/tex]
Differentiate y with respect to u:
[tex]\dfrac{\text{d}y}{\text{d}u}=7\cdot u^{7-1} \\\\\\ \dfrac{\text{d}y}{\text{d}u}=7u^6[/tex]
Now, differentiate u with respect to x using the quotient rule.
[tex]\boxed{\begin{array}{c}\underline{\textsf{Quotient Rule for Differentiation}}\\\\\dfrac{\text{d}}{\text{d}x}\left(\dfrac{\text{f}(x)}{\text{g}(x)}\right)=\dfrac{\text{g}(x)\text{f}\:'(x)-\text{f}(x)\text{g}\:'(x)}{\left(\text{g}(x)\right)^2}\end{array}}[/tex]
First, identify f(x) and g(x) and differentiate them separately:
[tex]\text{f}(x)=e^x \implies \text{f}'(x)=e^x[/tex]
[tex]\text{g}(x)=x+2 \implies \text{g}'(x)=1[/tex]
Now, put everything into the quotient rule formula:
[tex]\dfrac{\text{d}u}{\text{d}x}=\dfrac{(x+2) \cdot e^x-e^x\cdot 1}{(x+2)^2} \\\\\\ \dfrac{\text{d}u}{\text{d}x}=\dfrac{xe^x+2e^x-e^x}{(x+2)^2} \\\\\\ \dfrac{\text{d}u}{\text{d}x}=\dfrac{xe^x+e^x}{(x+2)^2} \\\\\\ \dfrac{\text{d}u}{\text{d}x}=\dfrac{e^x(x+1)}{(x+2)^2}[/tex]
Substitute the expressions for dy/du and du/dx into the chain rule formula:
[tex]\dfrac{\text{d}y}{\text{d}x}=7u^6 \cdot \dfrac{e^x(x+1)}{(x+2)^2}[/tex]
Substitute back in u:
[tex]\dfrac{\text{d}y}{\text{d}x}=7\left(\dfrac{e^x}{x+2}\right)^6 \cdot \dfrac{e^x(x+1)}{(x+2)^2}[/tex]
Therefore, the derivative of the given equation is:
[tex]\dfrac{\text{d}y}{\text{d}x}=7\left(\dfrac{e^x}{x+2}\right)^6 \cdot \dfrac{e^x(x+1)}{(x+2)^2}[/tex]
[tex]\dotfill[/tex]
To express the derivative in the preferred form, begin by factoring out [tex]\frac{e^x}{x+2}[/tex] from the second fraction and combine factors:
[tex]\dfrac{\text{d}y}{\text{d}x}=7\left(\dfrac{e^x}{x+2}\right)^6 \cdot \left(\dfrac{e^x}{x+2}\right) \cdot \left(\dfrac{x+1}{x+2} \right) \\\\\\ \dfrac{\text{d}y}{\text{d}x}=7\left(\dfrac{e^x}{x+2}\right)^7 \cdot \left(\dfrac{x+1}{x+2} \right)[/tex]
Now, multiply the second fraction by the constant 7:
[tex]\dfrac{\text{d}y}{\text{d}x}=\left(\dfrac{e^x}{x+2}\right)^7 \cdot \left(\dfrac{7x+7}{x+2} \right) \\\\\\[/tex]
Write the numerator of the second fraction as 7x + 14 - 7:
[tex]\dfrac{\text{d}y}{\text{d}x}=\left(\dfrac{e^x}{x+2}\right)^7 \cdot \left(\dfrac{7x+14-7}{x+2} \right)[/tex]
Factor out x + 2 from the first two terms of the numerator of the second fraction:
[tex]\dfrac{\text{d}y}{\text{d}x}=\left(\dfrac{e^x}{x+2}\right)^7 \cdot \left(\dfrac{7(x+2)-7}{x+2} \right) \\\\\\[/tex]
Split into two fractions:
[tex]\dfrac{\text{d}y}{\text{d}x}=\left(\dfrac{e^x}{x+2}\right)^7 \cdot \left(\dfrac{7(x+2)}{x+2} -\dfrac{7}{x+2}\right) \\\\\\[/tex]
Finally, cancel the common factor (x + 2):
[tex]\dfrac{\text{d}y}{\text{d}x}=\left(\dfrac{e^x}{x+2}\right)^7 \cdot \left(7 -\dfrac{7}{x+2}\right) \\\\\\[/tex]
Therefore, the derivative of the given equation in the preferred form is:
[tex]\dfrac{\text{d}y}{\text{d}x}=\left(\dfrac{e^x}{x+2}\right)^7\left(7 -\dfrac{7}{x+2}\right)[/tex]
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