Question

power of x is 4+x square +1 -------differentiate it w.r.t x

Answer 1

Answer on Question# #47295 – Mathematics – Differential Calculus | Equations

**Question:**

Power of $x$ is $4+x$ square $+1$ – differentiate it w.r.t $x$.

**Solution:**

$$y(x) = x^{(4+x)^2 + 1}.$$

Let us take the natural logarithm of left and right sides of this function:

$$ln y = ((4 + x)^2 + 1) \ln x.$$

Differentiating both sides, we have

$$\frac{y'}{y} = (2(4 + x)) \ln x + \frac{(4 + x)^2 + 1}{x},$$

where $y' = \frac{dy}{dx}$. Multiplying the expression (3) by the original function $y$, we finally obtain:

$$\begin{aligned} y' &= y \left[ (2(4 + x)) \ln x + \frac{(4 + x)^2 + 1}{x} \right] = \frac{x^{(4 + x)^2 + 1}}{x} \left[ (2(4 + x)) x \ln x + (4 + x)^2 + 1 \right] \\ &= x^{(4 + x)^2} (x^2 + 8x + 2(4 + x) x \ln x + 17) \end{aligned}$$

**Answer:** $y' = x^{(4 + x)^2} (x^2 + 8x + 2(4 + x) x \ln x + 17)$.

https://www.AssignmentExpert.com