To solve the derivative of In(sqrt x), the chain rule has to be used for this process. Learn the steps to solve the derivative of In(sqrt x) plus how to understand an application of this method in real life. Read Solving the Derivative of ln(sqrt x) Lesson Recommended...
Answer to: Find the first derivative of f(x) = \sqrt x \ln x. By signing up, you'll get thousands of step-by-step solutions to your homework...
Other differentiation rules will also be applied to get the derivative of the given function. Answer and Explanation:1 Applying the chain rule gives us: {eq}\begin{align*} \displaystyle f(x) & = \sqrt {\ln(x) + x}\f(x) & = (\ln(x) + x)^{\frac{1}{2}}\f'(x) & =......
dxd(ln(ex)) Simplifyln(ex):x =dxdx Apply the common derivative:dxdx=1=1 Enter your problem Popular Examples Frequently Asked Questions (FAQ) What is the derivative of ln(e^x) ? The derivative of ln(e^x) is 1 What is the first derivative of ln(e^x) ?
[1/1]Find the first derivative of function ln(x+sqrt(1+xx)) with respect to x: Solution:==Primitive function =ln(x+sqrt(xx+1))The first derivative function:d(ln(x+sqrt(xx+1)))dx(1+((xx((1)ln(x)+(x)(1)(x)))+0)∗12(xx+1)12)(x+sqrt(xx+1))xxln(x)2(x+sqrt(x...
1. Find the derivative of y = (\tan x)^{\ln x}. 2. Find the derivative of y = x^{\sqrt{2x. How do you find the 2nd derivative of y = (2x) / (x^2 - 4)? How do you find the derivative of y = log4(x)? How do you find the derivative of f(x) = (4x...
•csc(x)—cosecant •arcsec(x)—arcsecant •arccsc(x)—arccosecant •arsech(x)—inverse hyperbolic secant •arcsch(x)—inverse hyperbolic cosecant •|x|,abs(x)—absolute value •sqrt(x),root(x)—square root •exp(x)—e to the power of x ...
$\tiny{242.2q.3}$ $\textsf{find the derivative}\\$ \begin{align} \displaystyle y&=\frac{x \, \sqrt[]{x^2+1}}{(x+1)^{2/3}} \\ \ln{y}&=\ln x + \frac{1}{2}\ln(x^2+1) - \frac{2}{3}\ln(x+1)\\ \end{align} $\textit{thot this would help but what next?
Derivative $f’$ of function $f(x)=\arccos{x}$ is: \[\forall x \in ]–1, 1[ ,\quad f'(x) = -\frac{1}{\sqrt{1-x^2}}\] Proof Remember that function $\arcsin$ is the inverse function of $\cos$ : \[\left(f^{-1} \circ f\right)=\left(\cos \circ \arccos\right)...
Derivative $f’$ of function $f(x)=\arcsin{x}$ is: \(\forall x \in ]–1, 1[ ,\quad f'(x) = \dfrac{1}{\sqrt{1-x^2}}\) Proof Remember that function $\arcsin$ is the inverse function of $\sin$ : \[\left(f^{-1} \circ f\right)=\left(\sin \circ \arcsin\right)(...