Question

If y = x^(2x+3), then find dy dx at x=1.​

Answer 1

Answer:

$\frac{dy}{dx}$ at x = 1 for the given equation is 5.

Step-by-step explanation:

To find  $\frac{dy}{dx}$  at x = 1 for the given equation  $y = x^{(2x+3)}$,  we can use the logarithmic differentiation technique.

Let's start by taking the natural logarithm (ln) of both sides of the equation:

$ln(y) = ln(x^{(2x+3)} )$

Now, we can apply logarithmic rules to simplify the equation:

$ln(y) = (2x+3) * ln(x)$

Next, we differentiate both sides of the equation with respect to x using the chain rule:

$\frac{1}{y} * \frac{dy}{dx} = (2x+3) * (\frac{1}{x} ) + ln(x) * 2$

Simplifying further:

$\frac{dy}{dx} = y * [ \frac{ (2x+3)}{x} + 2ln(x)]$

Now, we substitute x = 1 into the equation to find dy/dx at x = 1:

$\frac{dy}{dx} = y * [ \frac{ (2(1)+3)}{1} + 2ln(1)]$

Since ln(1) = 0, the equation simplifies to:

$\frac{dy}{dx} = y * (\frac{5}{1} )$

Finally, substituting x+3) into the equation:

$\frac{dy}{dx} = x^{(2x+3)}* (\frac{5}{1} )$

At x = 1, the expression becomes:

$\frac{dy}{dx} = 1^{(2(1)+3)}* (\frac{5}{1} )$

$\frac{dy}{dx} = 1^5 * 5$

$\frac{dy}{dx} = 5$

Therefore, $\frac{dy}{dx}$ at x = 1 for the given equation is 5.

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