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Sagot :
Sure, let's solve the problem step-by-step.
Given the function:
[tex]\[ y = \ln\left(e^{3x^2} - \cos^9(x^2 + 3)\right) \][/tex]
We need to find the derivative \(\frac{dy}{dx}\).
Step 1: Identify the outer function and the inner function.
[tex]\[ u = e^{3x^2} - \cos^9(x^2 + 3) \][/tex]
[tex]\[ y = \ln(u) \][/tex]
Step 2: Use the chain rule of differentiation.
[tex]\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \][/tex]
Since \( y = \ln(u) \),
[tex]\[ \frac{dy}{du} = \frac{1}{u} \][/tex]
Step 3: Differentiate the inner function \( u \) with respect to \( x \).
[tex]\[ u = e^{3x^2} - \cos^9(x^2 + 3) \][/tex]
[tex]\[ \frac{du}{dx} = \frac{d}{dx}\left(e^{3x^2}\right) - \frac{d}{dx}\left(\cos^9(x^2 + 3)\right) \][/tex]
Step 4: Differentiate \( e^{3x^2} \) with respect to \( x \).
[tex]\[ \frac{d}{dx}\left(e^{3x^2}\right) = e^{3x^2} \cdot \frac{d}{dx}(3x^2) \][/tex]
[tex]\[ \frac{d}{dx}(3x^2) = 6x \][/tex]
So,
[tex]\[ \frac{d}{dx}\left(e^{3x^2}\right) = 6x \cdot e^{3x^2} \][/tex]
Step 5: Differentiate \( \cos^9(x^2 + 3) \) with respect to \( x \).
Using the chain rule again, let \( v = \cos(x^2 + 3) \), so \(\cos^9(x^2 + 3) = v^9\).
[tex]\[ \frac{d}{dx}\left(\cos^9(x^2 + 3)\right) = 9\cos^8(x^2 + 3) \cdot \frac{d}{dx}\left(\cos(x^2 + 3)\right) \][/tex]
Now, differentiate \( \cos(x^2 + 3) \):
[tex]\[ \frac{d}{dx}\left(\cos(x^2 + 3)\right) = -\sin(x^2 + 3) \cdot \frac{d}{dx}(x^2 + 3) \][/tex]
[tex]\[ \frac{d}{dx}(x^2 + 3) = 2x \][/tex]
So,
[tex]\[ \frac{d}{dx}\left(\cos(x^2 + 3)\right) = -2x \sin(x^2 + 3) \][/tex]
[tex]\[ \frac{d}{dx}\left(\cos^9(x^2 + 3)\right) = 9 \cdot \cos^8(x^2 + 3) \cdot (-2x \sin(x^2 + 3)) \][/tex]
[tex]\[ = -18x \cos^8(x^2 + 3) \sin(x^2 + 3) \][/tex]
Step 6: Combine the results from Steps 4 and 5.
[tex]\[ \frac{du}{dx} = 6x \cdot e^{3x^2} - 18x \cos^8(x^2 + 3) \sin(x^2 + 3) \][/tex]
Step 7: Substitute back into the chain rule expression.
[tex]\[ \frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} \][/tex]
[tex]\[ = \frac{1}{e^{3x^2} - \cos^9(x^2 + 3)} \cdot (6x e^{3x^2} - 18x \cos^8(x^2 + 3) \sin(x^2 + 3)) \][/tex]
[tex]\[ = \frac{6x e^{3x^2} - 18x \cos^8(x^2 + 3) \sin(x^2 + 3)}{e^{3x^2} - \cos^9(x^2 + 3)} \][/tex]
Therefore, the derivative is:
[tex]\[ \frac{dy}{dx} = \frac{6x e^{3x^2} + 18x \sin(x^2 + 3) \cos^8(x^2 + 3)}{e^{3x^2} - \cos^9(x^2 + 3)} \][/tex]
Given the function:
[tex]\[ y = \ln\left(e^{3x^2} - \cos^9(x^2 + 3)\right) \][/tex]
We need to find the derivative \(\frac{dy}{dx}\).
Step 1: Identify the outer function and the inner function.
[tex]\[ u = e^{3x^2} - \cos^9(x^2 + 3) \][/tex]
[tex]\[ y = \ln(u) \][/tex]
Step 2: Use the chain rule of differentiation.
[tex]\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \][/tex]
Since \( y = \ln(u) \),
[tex]\[ \frac{dy}{du} = \frac{1}{u} \][/tex]
Step 3: Differentiate the inner function \( u \) with respect to \( x \).
[tex]\[ u = e^{3x^2} - \cos^9(x^2 + 3) \][/tex]
[tex]\[ \frac{du}{dx} = \frac{d}{dx}\left(e^{3x^2}\right) - \frac{d}{dx}\left(\cos^9(x^2 + 3)\right) \][/tex]
Step 4: Differentiate \( e^{3x^2} \) with respect to \( x \).
[tex]\[ \frac{d}{dx}\left(e^{3x^2}\right) = e^{3x^2} \cdot \frac{d}{dx}(3x^2) \][/tex]
[tex]\[ \frac{d}{dx}(3x^2) = 6x \][/tex]
So,
[tex]\[ \frac{d}{dx}\left(e^{3x^2}\right) = 6x \cdot e^{3x^2} \][/tex]
Step 5: Differentiate \( \cos^9(x^2 + 3) \) with respect to \( x \).
Using the chain rule again, let \( v = \cos(x^2 + 3) \), so \(\cos^9(x^2 + 3) = v^9\).
[tex]\[ \frac{d}{dx}\left(\cos^9(x^2 + 3)\right) = 9\cos^8(x^2 + 3) \cdot \frac{d}{dx}\left(\cos(x^2 + 3)\right) \][/tex]
Now, differentiate \( \cos(x^2 + 3) \):
[tex]\[ \frac{d}{dx}\left(\cos(x^2 + 3)\right) = -\sin(x^2 + 3) \cdot \frac{d}{dx}(x^2 + 3) \][/tex]
[tex]\[ \frac{d}{dx}(x^2 + 3) = 2x \][/tex]
So,
[tex]\[ \frac{d}{dx}\left(\cos(x^2 + 3)\right) = -2x \sin(x^2 + 3) \][/tex]
[tex]\[ \frac{d}{dx}\left(\cos^9(x^2 + 3)\right) = 9 \cdot \cos^8(x^2 + 3) \cdot (-2x \sin(x^2 + 3)) \][/tex]
[tex]\[ = -18x \cos^8(x^2 + 3) \sin(x^2 + 3) \][/tex]
Step 6: Combine the results from Steps 4 and 5.
[tex]\[ \frac{du}{dx} = 6x \cdot e^{3x^2} - 18x \cos^8(x^2 + 3) \sin(x^2 + 3) \][/tex]
Step 7: Substitute back into the chain rule expression.
[tex]\[ \frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} \][/tex]
[tex]\[ = \frac{1}{e^{3x^2} - \cos^9(x^2 + 3)} \cdot (6x e^{3x^2} - 18x \cos^8(x^2 + 3) \sin(x^2 + 3)) \][/tex]
[tex]\[ = \frac{6x e^{3x^2} - 18x \cos^8(x^2 + 3) \sin(x^2 + 3)}{e^{3x^2} - \cos^9(x^2 + 3)} \][/tex]
Therefore, the derivative is:
[tex]\[ \frac{dy}{dx} = \frac{6x e^{3x^2} + 18x \sin(x^2 + 3) \cos^8(x^2 + 3)}{e^{3x^2} - \cos^9(x^2 + 3)} \][/tex]
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