What is the derivative of ${{2}^{x}}$?
Answer
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Hint: To find the derivative of ${{2}^{x}}$, we are going to first of all equate this ${{2}^{x}}$ to y and then we take ${{\log }_{e}}$ or $\ln $ on both the sides of the equation. Then we are going to use the property of the logarithm which states that $\log {{a}^{b}}=b\log a$. Then we will take the derivative on both the sides of log expression and y.
Complete step-by-step solution:
The function in x which we are going to derivate is as follows:
${{2}^{x}}$
Equating the above expression to y we get,
$y={{2}^{x}}$ ………….. (1)
Taking $\ln $ on both the sides of the above equation we get,
$\ln y=\ln {{2}^{x}}$ …………. (2)
We know the property of logarithm which states that:
$\log {{a}^{b}}=b\log a$
Using the above property in the R.H.S of eq. (2) we get,
$\ln y=x\ln 2$ ………….. (3)
Taking derivative with respect to x on both the sides of the above equation we get,
We know the derivative of $\ln x$ with respect to x is equal to $\dfrac{1}{x}$. Writing this derivative expression mathematically we get,
$\dfrac{d\ln x}{dx}=\dfrac{1}{x}$
Also, we know the derivative of x with respect to x is 1. The mathematical expression for this derivative is as follows:
$\dfrac{dx}{dx}=1$
Using the above derivatives and applying them in the eq. (3) we get,
$\dfrac{1}{y}\dfrac{dy}{dx}=\ln 2$
Now, multiplying y on both the sides of the above equation we get,
$\dfrac{dy}{dx}=y\ln 2$
Substituting the value of y from eq. (1) in the R.H.S of the above equation we get,
$\dfrac{dy}{dx}={{2}^{x}}\ln 2$
Hence, we have calculated the derivative of ${{2}^{x}}$ as ${{2}^{x}}\ln 2$.
Note: The alternative approach to solve the above problem is that, we know the derivative of ${{a}^{x}}$ with respect to x which is equal to:
$\dfrac{d{{a}^{x}}}{dx}=\left( \ln a \right){{a}^{x}}$
Applying the above derivative rule in the derivative of ${{2}^{x}}$ with respect to x then “a” becomes 2 in the above differentiation and we get,
$\dfrac{d\left( {{2}^{x}} \right)}{dx}=\left( \ln 2 \right){{2}^{x}}$
Complete step-by-step solution:
The function in x which we are going to derivate is as follows:
${{2}^{x}}$
Equating the above expression to y we get,
$y={{2}^{x}}$ ………….. (1)
Taking $\ln $ on both the sides of the above equation we get,
$\ln y=\ln {{2}^{x}}$ …………. (2)
We know the property of logarithm which states that:
$\log {{a}^{b}}=b\log a$
Using the above property in the R.H.S of eq. (2) we get,
$\ln y=x\ln 2$ ………….. (3)
Taking derivative with respect to x on both the sides of the above equation we get,
We know the derivative of $\ln x$ with respect to x is equal to $\dfrac{1}{x}$. Writing this derivative expression mathematically we get,
$\dfrac{d\ln x}{dx}=\dfrac{1}{x}$
Also, we know the derivative of x with respect to x is 1. The mathematical expression for this derivative is as follows:
$\dfrac{dx}{dx}=1$
Using the above derivatives and applying them in the eq. (3) we get,
$\dfrac{1}{y}\dfrac{dy}{dx}=\ln 2$
Now, multiplying y on both the sides of the above equation we get,
$\dfrac{dy}{dx}=y\ln 2$
Substituting the value of y from eq. (1) in the R.H.S of the above equation we get,
$\dfrac{dy}{dx}={{2}^{x}}\ln 2$
Hence, we have calculated the derivative of ${{2}^{x}}$ as ${{2}^{x}}\ln 2$.
Note: The alternative approach to solve the above problem is that, we know the derivative of ${{a}^{x}}$ with respect to x which is equal to:
$\dfrac{d{{a}^{x}}}{dx}=\left( \ln a \right){{a}^{x}}$
Applying the above derivative rule in the derivative of ${{2}^{x}}$ with respect to x then “a” becomes 2 in the above differentiation and we get,
$\dfrac{d\left( {{2}^{x}} \right)}{dx}=\left( \ln 2 \right){{2}^{x}}$
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