Answer :
Final answer:
The derivative of y = e⁽⁷ˣ⁷⁺³ˣ³⁺¹⁾ + log₅ (x⁵ + 5x + 5) is y' = (7x⁷ + 3x³ + 1) * e⁽⁷ˣ⁷⁺³ˣ³⁺¹⁾ + (1/ln(5)) * [5x⁴/(x⁵ + 5x + 5)].
Explanation:
To find the derivative of the given function, we will use the chain rule and logarithmic differentiation rule.
Let's start by finding the derivative of the exponential function, e⁽⁷ˣ⁷⁺³ˣ³⁺¹⁾. Using the chain rule, we have:
d/dx(e⁽⁷ˣ⁷⁺³ˣ³⁺¹⁾) = (7x⁷ + 3x³ + 1) * e⁽⁷ˣ⁷⁺³ˣ³⁺¹⁾
Next, let's find the derivative of the logarithmic function, log₅ (x⁵ + 5x + 5). Using the logarithmic differentiation rule, we have:
d/dx(log₅ (x⁵ + 5x + 5)) = (1/ln(5)) * [5x⁴/(x⁵ + 5x + 5)]
Now, we can find the derivative of the given function, y = e⁽⁷ˣ⁷⁺³ˣ³⁺¹⁾ + log₅ (x⁵ + 5x + 5), by adding the derivatives of the exponential and logarithmic functions:
y' = (7x⁷ + 3x³ + 1) * e⁽⁷ˣ⁷⁺³ˣ³⁺¹⁾ + (1/ln(5)) * [5x⁴/(x⁵ + 5x + 5)]
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