Need help in some questions

[Ali]

Helllo My dearest friends i need some help can you guys solve these question with respect to differential

[Guest]

Here is a free online tool that calculate derivatives step by step
https://www.math10.com/en/problem-solve ... olver.html

Here is the first one:
https://www.math10.com/en/problem-solve ... qrt%28x%29

[Guest]

There doesn't seem to be any statement as to what the question is!
We are told that "d/dx= differential" but there is no "d/dx" anywhere else in the problem!

I am guessing that the problem is to find the derivative of all these functions.

Almost the first thing one learns in an introductor Calculus class is that the derivative of [tex]x^n[/tex], where n can be any number, with respect to x, is [tex]nx^{n-1}[/tex] and most of these problems are of that form.
1: [tex]\sqrt{x}= x^{1/2}[/tex] so its derivative is [tex]\frac{1}{2}x^{1/2- 1}= \frac{1}{2}x^{-1/2]= \frac{1}[2\sqrt{x}}[/tex].

2: [tex]x^{3/2}[/tex] so its derivative is [tex]\frac{3}{2}x^{3/2- 1}= \frac{3}{2}x^{1/2}[/tex].

3: [tex]4x= 4x^1[/tex] so its derivative is [tex]4(1)x^{1- 1}= 4x^0= 4[/tex].
(But even before that rule one usually learns that the derivative of a linear function, ax+ b, is a constant, the slope, a. Here we have 4x+0 so the slope is 4.

For 4 and many after, we need the "chain rule": if y is a function of some variable, u, and u is itelf a function of x, then y can be written as a function of x and [tex]\frac{dy}{dx}= \frac{dy}{du}\frac{du}{dx}[/tex]

4:[tex](x^2+ 1)^2[/tex].
Let [tex]u= x^2+ 2[/tex] so that [tex]y= u^2[/tex]. Then [tex]\frac{dy}{du}= 2u[/tex] and [tex]\frac{du}{dx}= 2x[/tex] so [tex]\frac{dy}{dx}= 2u(2x)= 4(x^2+ 2)x= 4x^3+ 8x[/tex].
Of course we could simply multply this out to get [tex](x^2+ 2)^2= x^4+ 4x^2+ 4[/tex] and the derivative is [tex]4x^3+ 4(2x^1)= 4x^3+ 8x[/tex].

The others are much the same though 7 and 8 appear to be functions of other variables that x.
7 has [tex]y^2[/tex] and 8 has [tex]t^2[/tex] so you will need to differentiate with respect to y and t rather than x. Of course, the derivative is the same except that you use y and t rather than x. The derivative of [tex]y^2[/tex] with respect to y is [tex]2y[/tex] and the derivative of [tex]t^2[/tex] with respect to t is [tex]2t[/tex].

IF the instructions specifically say to differentiate with respect to x then you need to use the chain rule again. The derivative of [tex]y^2[/tex] with respect to x is [tex]\frac{dy^2}{dy}\frac{dy}{dx}= 2y\frac{dy}{dx}[/tex] and since we do not know how y is a function of x, all we can do is leave it like that,
[tex]2y\frac{dy}{dx}[/tex].