Question

What is the derivative of sin(ln x) + cos(ln x) with respect to x at x = e?

1. \(\dfrac{cos 1-sin1}{e}\)
2. \(\dfrac{sin1-cos1}{e}\)
3. \(\dfrac{cos1+sin1}{e}\)
4. 1

Answer 1

Correct Answer - Option 1 : \(\dfrac{cos 1-sin1}{e}\)

CONCEPT:

• \(\frac{{d\left( {\sin x} \right)}}{{dx}} = \cos x\)
• \(\frac{{d\left( {\cos x} \right)}}{{dx}} = \; - \sin x\)
• \(\frac{{d\left( {\ln x} \right)}}{{dx}} = \frac{1}{x}\)

CALCULATION:

Let f(x) = sin(ln x) + cos(ln x)

As we know that, \(\frac{{d\left( {\sin x} \right)}}{{dx}} = \cos x, \frac{{d\left( {\cos x} \right)}}{{dx}} = \; - \sin x \ and \ \frac{{d\left( {\ln x} \right)}}{{dx}} = \frac{1}{x}\)

⇒ \(f'\left( x \right) = \frac{{\cos \left( {\ln x} \right)}}{x} - \frac{{\sin \left( {\ln x} \right)}}{x}\)

⇒ \(f'\left( e \right) = \frac{{\cos \left( {\ln e} \right)}}{e} - \frac{{\sin \left( {\ln e} \right)}}{e}\)

As we know that, ln(e) = 1

⇒ \(f'(e) = \frac{cos1 - sin1}{e}\)

Hence, option 1 is correct.