Mustapha Mallum
more than 4 weeks ago
Differentiate y=loga^x ?
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more than 4 weeks ago
The given function is y = loga^x.
The logarithm function is the inverse of the exponential function. In this case, the base of the logarithm is "a" and the exponent is "x".
To differentiate this function, we can use the chain rule. The derivative of loga^x with respect to x is:
dy/dx = (1/(x * ln(a))) * (d/dx) (a^x)
Using the chain rule, the derivative of a^x with respect to x is:
d/dx (a^x) = ln(a) * a^x
Therefore, the derivative of y = loga^x with respect to x is:
dy/dx = (1/(x * ln(a))) * (ln(a) * a^x)
dy/dx = a^x / x
The logarithm function is the inverse of the exponential function. In this case, the base of the logarithm is "a" and the exponent is "x".
To differentiate this function, we can use the chain rule. The derivative of loga^x with respect to x is:
dy/dx = (1/(x * ln(a))) * (d/dx) (a^x)
Using the chain rule, the derivative of a^x with respect to x is:
d/dx (a^x) = ln(a) * a^x
Therefore, the derivative of y = loga^x with respect to x is:
dy/dx = (1/(x * ln(a))) * (ln(a) * a^x)
dy/dx = a^x / x
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