Question

derivative. of log x3/(1+2x) with respect to x at x = 1​

Answer 1

Answer:

Let, y=x

2

and z=logx. We are to find

dz

dy

.

This gives

dx

dy

=2x and

dx

dz

=

x

1

.

Now

dz

dy

=

dx

dz

dx

dy

=2x

2

.

Step-by-step explanation:

I hope this helps you dear

Answer 2

Question: Find the derivative of $log\left(\frac{x^3}{1+2x}\right)$ with respect to $x$ at $x=1$.

Answer:

The derivative of $log\left(\frac{x^3}{1+2x}\right)$ with respect to $x$ at $x=1$ is $7/3$.

Step-by-step explanation:

Step 1 of 3

Consider the function as follows:

$y=log\left(\frac{x^3}{1+2x}\right)$

Differentiate both the sides with respect to $x$ as follows:

$\frac{dy}{dx} =\frac{d}{dx}\left[log\left(\frac{x^3}{1+2x}\right)\right]$

Differentiate the logarithm function $logx^3$ using chain rule.

$\frac{dy}{dx} =\frac{1}{\frac{x^3}{1+2x}} \cdot \frac{d}{dx} \left(\frac{x^3}{1+2x}\right)$

As derivative of log$x$ is $1/x$, i.e., $\frac{d}{dx}(logx)=\frac{1}{x}$.

Further, simplify as follows:

$\frac{dy}{dx} =\frac{1+2x}{x^3} \cdot \frac{d}{dx} \left(\frac{x^3}{1+2x}\right)$

Step 2 of 3

Using quotient rule, differentiate the right-hand side.

$\frac{dy}{dx} = \frac{1+2x}{x^3}\cdot \frac{(1+2x)\frac{d}{dx}(x^3)-(x^3) \frac{d}{dx}(1+2x)}{(1+2x)^2}$

$\frac{dy}{dx} = \frac{1+2x}{x^3}\cdot \frac{(1+2x)3x^2-(x^3)2}{(1+2x)^2}$

On simplifying, we get

$\frac{dy}{dx} = \frac{1}{x^3(1+2x)}\cdot(3x^2+6x^3-2x^3)$

$\frac{dy}{dx} = \frac{3+4x}{x(1+2x)}$ . . . . . (1)

Step 3 of 3

Finding the value of derivative at $x=1$.

Substitute the value $1$ for $x$ in the equation (1) as follows:

$\frac{dy}{dx}|_{x=1}=\frac{3+4(1)}{1(1+2(1))}$

Now, simplify the equation.

⇒ $\frac{dy}{dx}|_{x=1}=\frac{3+4}{(1+2)}$

               $=\frac{7}{3}$

Therefore, the derivative of $log\left(\frac{x^3}{1+2x}\right)$ with respect to $x$ at $x=1$ is $7/3$.

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