pls integrate and differentiate e^x^2
and x/y square root of x^2+y^2 in terms of x.
To integrate and differentiate the given expressions, let's start with each one separately:
1. Integrating e^x^2:
To integrate e^x^2, we can use the technique of substitution. Let's denote x^2 as t, then dx = 2x dt. Rearranging, we have dt = dx / (2x).
So, the integral of e^x^2 becomes:
∫ e^x^2 dx = ∫ e^t * (dx / (2x))
= (1/2) ∫ e^t * (1/x) dx
= (1/2) ∫ e^t / x dx
Now, we can integrate ∫ e^t / x dx:
Let's assume that ∫ e^t / x dx = F(t).
Applying the technique of integration by parts, we have:
F(t) = ∫ e^t / x dx
= e^t * ln(x) - ∫ (d/dt[e^t]) * ln(x) dx
= e^t * ln(x) - ∫ (e^t / x) dx
= e^t * ln(x) - F(t)
Rearranging, we have:
2F(t) = e^t * ln(x) + C
F(t) = (1/2) * e^t * ln(x) + C
Since t = x^2, the final result becomes:
∫ e^x^2 dx = (1/2) * e^(x^2) * ln(x) + C
2. Differentiating (x/y) * sqrt(x^2 + y^2) in terms of x:
Let's denote the given expression as f(x).
f(x) = (x/y) * sqrt(x^2 + y^2)
To differentiate f(x) with respect to x, we can apply the product rule and the chain rule:
f'(x) = (1/y) * sqrt(x^2 + y^2) + (x/y^2) * (1/2) * (x^2 + y^2)^(-1/2) * 2x
Simplifying this expression, we have:
f'(x) = (1/y) * sqrt(x^2 + y^2) + (x^2/ y^2) * (x^2 + y^2)^(-1/2)
Therefore, the derivative of (x/y) * sqrt(x^2 + y^2) with respect to x is:
f'(x) = (1/y) * sqrt(x^2 + y^2) + (x^2/ y^2) * (x^2 + y^2)^(-1/2)
I hope this helps! Let me know if you have any further questions.