Find the derivative of ${(\ln x)^2}$.
Answer
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Hint: Derivative of the function can be simply defined as the Slope of the curve. Derivative of a function is an increasing or decreasing function, depending upon the value x.
Derivative of the function f(x) = $\mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{f({x_0} + \Delta x) - f({x_0})}}{{\Delta x}}$ in the mathematical terms. Derivatives of the function are generated by this formula, but there is no need to find derivatives of the general functions, whose derivative is already known to us, we just need to remember them and make use in finding derivatives of the complex equations.
Complete step-by-step solution:
Let us consider $y = {\left( {\ln x} \right)^2}$
To find: Derivative of y
Here, we make use of the chain rule to solve the derivatives. For that, we need to make certain adjustments and we make them by substitutions.
Let $u = \left ({\ln x} \right)$
Differentiate u with respect to x, we get
$\dfrac {d} {{dx}} u = \dfrac{d}{{dx}}\ln x$
Derivative of $\ln x$is $\dfrac {1} {x}$
Therefore, $\dfrac{d}{{dx}}u = \dfrac{d}{{dx}}\ln x = \dfrac{1}{x}$ ------$\left(1\right)$
Substituting u in the original function y, we get
$y = {u^2}$
When we differentiate $y = {\left( {\ln x} \right)^2}$, with respect to u, we get
$\dfrac{d}{{du}}y = \dfrac{d}{{du}}{u^2}$
Derivative of ${x^n}$ is $n \times {x^{n - 1}}$, this is the general form.
Here \[x = u\] and \[n = 2\], therefore, derivative of ${u^2}$ becomes $2 \times {u^{2 - 1}}$, i.e. we get
$\dfrac{d}{{du}}y = 2u$-------$\left(1\right)$
Using chain rule, i.e. $\dfrac{d}{{dx}}y = \dfrac{d}{{du}}{{u}^{2}}\times \dfrac{d}{{dx}}u$, we get
$\dfrac{d}{{dx}}y = 2u \times \dfrac{1}{x}$ --- By using equations 1 and 2
Substituting the value of u in the above equation, we get
$\dfrac{d}{{dx}}y = 2\ln x \times \dfrac{1}{x}$
Therefore, derivative of ${\left( {\ln x} \right)^2}$ is $\dfrac{{2\ln x}}{x}$
Note: Derivatives of standard functions must be known, making you solve the sum easily and quickly. Also, substitutions should be done properly. Mostly try to solve derivatives of complex functions by substitutions and later using chain rule.
Derivative of the function f(x) = $\mathop {\lim }\limits_{\Delta x \to 0} \dfrac{{f({x_0} + \Delta x) - f({x_0})}}{{\Delta x}}$ in the mathematical terms. Derivatives of the function are generated by this formula, but there is no need to find derivatives of the general functions, whose derivative is already known to us, we just need to remember them and make use in finding derivatives of the complex equations.
Complete step-by-step solution:
Let us consider $y = {\left( {\ln x} \right)^2}$
To find: Derivative of y
Here, we make use of the chain rule to solve the derivatives. For that, we need to make certain adjustments and we make them by substitutions.
Let $u = \left ({\ln x} \right)$
Differentiate u with respect to x, we get
$\dfrac {d} {{dx}} u = \dfrac{d}{{dx}}\ln x$
Derivative of $\ln x$is $\dfrac {1} {x}$
Therefore, $\dfrac{d}{{dx}}u = \dfrac{d}{{dx}}\ln x = \dfrac{1}{x}$ ------$\left(1\right)$
Substituting u in the original function y, we get
$y = {u^2}$
When we differentiate $y = {\left( {\ln x} \right)^2}$, with respect to u, we get
$\dfrac{d}{{du}}y = \dfrac{d}{{du}}{u^2}$
Derivative of ${x^n}$ is $n \times {x^{n - 1}}$, this is the general form.
Here \[x = u\] and \[n = 2\], therefore, derivative of ${u^2}$ becomes $2 \times {u^{2 - 1}}$, i.e. we get
$\dfrac{d}{{du}}y = 2u$-------$\left(1\right)$
Using chain rule, i.e. $\dfrac{d}{{dx}}y = \dfrac{d}{{du}}{{u}^{2}}\times \dfrac{d}{{dx}}u$, we get
$\dfrac{d}{{dx}}y = 2u \times \dfrac{1}{x}$ --- By using equations 1 and 2
Substituting the value of u in the above equation, we get
$\dfrac{d}{{dx}}y = 2\ln x \times \dfrac{1}{x}$
Therefore, derivative of ${\left( {\ln x} \right)^2}$ is $\dfrac{{2\ln x}}{x}$
Note: Derivatives of standard functions must be known, making you solve the sum easily and quickly. Also, substitutions should be done properly. Mostly try to solve derivatives of complex functions by substitutions and later using chain rule.
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