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?

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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:

\(\displaystyle y=f(x)+g(x)\)

What is \(\displaystyle \d{y}{x}\)?

 

[rahulk1]

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

 

[MarkFL]

Differentiation is linear, that is:

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

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:

\(\displaystyle f(x)=1+x+x^2+x^3+x^4\)

then:

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

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

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

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:

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

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:

\(\displaystyle u=\log(x)\)

Using the change of base formula, we may write:

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

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

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

For the third term, let's write:

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

Using the quotient rule, we obtain:

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

So, putting it all together, we have:

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