Given:
[tex]sinh(f(x))=1+x^2[/tex]
To find:
The value of f'(x).
Solution:
Formulae used:
[tex]\dfrac{d}{dx}sinh(x)=cosh(x)[/tex]
[tex]\dfrac{d}{dx}x^n=nx^{n-1}[/tex]
[tex]\dfrac{d}{dx}C=0[/tex]
Chain rule:
[tex]\dfrac{d}{dx}[f(g(x))]=f'(g(x))g'(x)[/tex]
Where C is an arbitrary constant.
We have,
[tex]sinh(f(x))=1+x^2[/tex]
Differentiate with respect to x.
[tex]cosh(f(x))f'(x)=0+2x[/tex]
[tex]f'(x)=\dfrac{2x}{cosh(f(x))}[/tex]
Therefore, the required values is [tex]f'(x)=\dfrac{2x}{cosh(f(x))}[/tex].
