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Find the derivative of the following function.

y=log_(3)(-8x^(6)-6x^(5))
Answer: 
y^(')=

Find the derivative of the following function.\newliney=log3(8x66x5) y=\log _{3}\left(-8 x^{6}-6 x^{5}\right) \newlineAnswer: y= y^{\prime}=

Full solution

Q. Find the derivative of the following function.\newliney=log3(8x66x5) y=\log _{3}\left(-8 x^{6}-6 x^{5}\right) \newlineAnswer: y= y^{\prime}=
  1. Identify Function & Derivative Type: Identify the function and the type of derivative to be found.\newlineWe need to find the derivative of the function y=log3(8x66x5)y = \log_3(-8x^6 - 6x^5) with respect to xx. This is a logarithmic differentiation problem where the base of the logarithm is 33.
  2. Apply Logarithmic Differentiation Rule: Apply the logarithmic differentiation rule.\newlineThe derivative of logb(u)\log_b(u) with respect to xx is (1u)(dudx)(1log(b))(\frac{1}{u}) \cdot (\frac{du}{dx}) \cdot (\frac{1}{\log(b)}), where uu is a function of xx and bb is the base of the logarithm. In our case, u=8x66x5u = -8x^6 - 6x^5 and b=3b = 3.
  3. Differentiate Inside Function: Differentiate the inside function u=8x66x5u = -8x^6 - 6x^5 with respect to xx. Using the power rule, the derivative of uu with respect to xx is dudx=8×6x616×5x51=48x530x4\frac{du}{dx} = -8 \times 6x^{6-1} - 6 \times 5x^{5-1} = -48x^5 - 30x^4.
  4. Substitute Derivative into Formula: Substitute the derivative of uu into the differentiation formula.\newlineNow we have dudx=48x530x4\frac{du}{dx} = -48x^5 - 30x^4, and we can substitute this into the formula from Step 22 to get the derivative of yy with respect to xx.\newliney=1(8x66x5)(48x530x4)1log(3)y' = \frac{1}{(-8x^6 - 6x^5)} \cdot (-48x^5 - 30x^4) \cdot \frac{1}{\log(3)}
  5. Simplify Expression: Simplify the expression.\newlineWe can simplify the expression by multiplying the terms together. However, we must be careful with the negative sign in the denominator.\newliney=48x530x4(8x66x5)log(3)y' = \frac{-48x^5 - 30x^4}{(-8x^6 - 6x^5) \cdot \log(3)}
  6. Factor Out Common Terms: Factor out common terms if possible.\newlineIn this case, we can factor out an x4x^4 from the numerator and denominator to simplify the expression further.\newliney=x4(48x30)x4(8x26x)log(3)y' = \frac{x^4(-48x - 30)}{x^4(-8x^2 - 6x) \cdot \log(3)}\newliney=48x30(8x26x)log(3)y' = \frac{-48x - 30}{(-8x^2 - 6x) \cdot \log(3)}
  7. Check for Simplifications: Check for any possible simplifications or cancellations. There are no further simplifications or cancellations that can be made in this expression. Therefore, the final derivative is: y=48x30(8x26x)log(3)y' = \frac{-48x - 30}{(-8x^2 - 6x) \cdot \log(3)}

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