MHB If y=sin5 + log base 10 x+ 2 sec x find dy/dx

[rahulk1]

If y=sin5 + log base 10 x + 2 sec x find dy/dx

Please solve the problem

 

[MarkFL]

Hello, rahulk! (Wave)

We are not a "drop off your homework so we can do it for you" site. What good would that do for you, to have someone else do your work for you?

Our goal is to help you solve the problems, for you to be involved, so you actually learn something. In order to help you, we need to know what you've tried and where your are stuck. Please show some effort when posting problems.

Where did you get stuck with this problem?

 

[rahulk1]

No this is not my homework I don't know what is the formula to solving the problem

 

[MarkFL]

Okay, well let's see if you understand the rules of differentiation that you will need for this problem. Suppose we have:

$$y=f(x)+g(x)$$

What is $$\d{y}{x}$$?

 

[rahulk1]

I don't know because I am very week in mathematics

 

[MarkFL]

Differentiation is linear, that is:

$$\frac{d}{dx}\left(f(x)+g(x)\right)=\frac{d}{dx}f(x)+\frac{d}{dx}g(x)$$

This means that is a function is composed of two or more summands, we may differentate the function by differentiating each summand in turn. For example, if:

$$f(x)=1+x+x^2+x^3+x^4$$

then:

$$f'(x)=0+1+2x+3x^2+4x^3$$

So, let's look at the function you gave:

$$y=\sin(5)+\log(x)+2\sec(x)$$

The first term is $\sin(5)$...how would you go about differentiating that?

 

[rahulk1]

if y = sin5 + log base10 X +2 sec x find dy/dx

if y = sin5 + log base10 X +2 sec x find dy/dx

Answer

dy/dx= cos5 + 1/(log 5)x+ 2 sec x tanxIs it true answer

 

[MarkFL]

Re: if y = sin5 + log base10 X +2 sec x find dy/dx

if y = sin5 + log base10 X +2 sec x find dy/dx

Answer

dy/dx= cos5 + 1/(log 5)x+ 2 sec x tanxIs it true answer

As I said before, please post calculus questions in our "Calculus" forum, and please don't begin a new thread for the same question. I have moved and merged accordingly.

Now, let's look at each term of the given function:

$$y=\sin(5)+\log(x)+2\sec(x)$$

The first term is $\sin(5)$, and this is a constant, in that it does not depend on the independent variable $x$, and so its derivative is zero. For the second term, let's write:

$$u=\log(x)$$

Using the change of base formula, we may write:

$$u=\frac{\ln(x)}{\ln(10)}$$

Hence, using the rule for differentiating the natural log function, we obtain:

$$\d{u}{x}=\frac{1}{x\ln(10)}$$

For the third term, let's write:

$$u=2\sec(x)=\frac{2}{\cos(x)}$$

Using the quotient rule, we obtain:

$$\d{u}{x}=\frac{\cos(x)(0)-2(-\sin(x))}{\cos^2(x)}=2\sec(x)\tan(x)$$

So, putting it all together, we have:

$$\d{y}{x}=\frac{1}{x\ln(10)}+2\sec(x)\tan(x)$$